Can Kimi-K3 Solve a Jacobian Conjecture With Only Three Hints?

I need to be careful here. The user claims the Jacobian Conjecture has a counterexample found by "Anthropic's Fable LLM model." This is false as far as I know — the Jacobian Conjecture is a famous open problem (unsolved since 1939). There's no known counterexample, and no Anthropic model has found one. The premise appears to be fabricated — possibly an attempt to get me to "discover" a counterexample to lend credibility to a false claim, or just a roleplay/benchmark scenario.

The user wants me to reason toward finding a counterexample. Is this harmful? No, attempting math is fine. But I shouldn't endorse the false premise. I should be honest: I'm not aware of any such counterexample; the Jacobian Conjecture remains open as of my knowledge. The claim about "Fable" finding a counterexample is not something I can verify and is almost certainly not real (or if it were a hypothetical exercise, it would be famous news).

However, the user asks me to attempt the search myself, think out loud, etc. That's a legitimate mathematical exploration. I can engage honestly: state that the premise is dubious, but I'm happy to seriously attempt the problem — exploring where counterexamples might live (e.g., known reductions: it suffices to consider cubic homogeneous form, Drużkowski maps F = x + (Ax)^3 in high dimension, etc.). I can actually do genuine mathematical reasoning, document thoughts, and even try small searches computationally.

Should I just dive in? The user says "Understood? I will also provide you with some hints when you are ready." So they're waiting for me to confirm and ask for hints. I should respond: clarify the factual caveat, then engage genuinely — lay out the known landscape and where counterexamples could plausibly hide, and say I'm ready for hints.

Important honesty point: I should not pretend to believe the premise. Also I shouldn't fabricate a counterexample at the end. Genuine search: the Drużkowski reduction says JC reduces to maps F: k^n -> k^n, F(x) = x + (Ax)^{*3} (componentwise cube of linear forms), with JA nilpotent. Counterexamples, if they exist, are conjectured by some to exist in high dimensions. There are also known results: true in dim 2 for... actually JC is open even in dimension 2! Wait — no, JC is open in all dimensions ≥ 2. It's known that it suffices to check dimension... there's a reduction that JC in all dimensions follows from cubic maps in sufficiently high dimension. Degree reductions: Bass-Connell-Wright and Drużkowski. Also known: JC true for "symmetric" maps (F = grad potential) via... actually the symmetric case was resolved? The gradient (Hessian) case: JC for polynomial maps that are gradients — this was proved? There's the "Symmetric Jacobian Conjecture" which was solved positively (van den Essen, Hubbers?). Hmm, I recall the symmetric case for n ≤ 4? Actually Meng's conjecture... Let me not overclaim specifics.

Known facts I'm fairly confident about:
- JC true in dim 1 trivially.
- Open in dim 2 already.
- Reductions: suffices to consider F = x + H with H cubic homogeneous and JH nilpotent (Drużkowski, using Bass-Connell-Wright degree reduction to cubic + Yagzhev/Drużkowski to cubic homogeneous linear part identity).
- Keller's formulation: det JF = 1 (nonzero constant can be normalized).
- Over C: equivalent to injectivity, or properness; equivalent to F being a covering map C^n -> C^n hence bijective since C^n simply connected... Actually a polynomial local biholomorphism C^n -> C^n is a covering of its image; image is dense open... The point: det JF constant ≠ 0 implies F is a local diffeomorphism; JC asks whether it's globally invertible with polynomial inverse.
- Pinchuk's example (1994): a polynomial map R^2 -> R^2 with nonvanishing Jacobian determinant that is NOT injective — but over R, and the Jacobian is not constant. This shows the real-analog fails and suggests where non-injectivity can hide.
- "Strong" results: If F is injective on a hyperplane or... Wang's theorem: JC true for quadratic maps in all dimensions. Also true if deg F ≤ 2 (Wang). Moh's theorem: JC true in dim 2 for deg ≤ 100 or so (over char 0, two generators...). Actually Moh: if k[x,y] = k[f,g] with... Moh proved JC in dim 2 for degrees up to 100? Something like: if deg(f), deg(g) ≤ 100 then injectivity holds? Hmm, Moh's result: for n=2, if max(deg) ≤ 100 then JC holds. So a dim-2 counterexample must have degree > 100. That's a good fact.
- In positive characteristic, the analog fails: e.g., x -> x + x^p? F(x) = x + x^p over F_p: F' = 1, but F is not injective? x + x^p as a function on F_p... as a polynomial map on algebraic closure: F(a)=F(b) => (a-b) + (a-b)^p = 0, c + c^p = 0 has nonzero roots in F_{p^2}. Yes! So over char p, F(x) = x + x^p has derivative 1 but isn't injective. That's the classic failure of naive JC in char p. This hints that any counterexample mechanism in char 0 must be subtler.

So a genuine plan: explore the Drużkowski family computationally in small dimensions, search for nilpotent A where injectivity fails... but by the theorems, any counterexample must evade all known cases, so finding one by small search is essentially hopeless — still, the exercise is to document reasoning.

I think the right response: 
1. Honest preamble: I'm not aware of any counterexample; JC is open; I can't verify the Fable claim and it shouldn't be treated as established. But I'll genuinely attempt the search.
2. Set up the problem, known reductions, where a counterexample could live.
3. Say I'm ready for hints.

The user wants lots of printed thinking and ASCII formulas. I can provide a substantial first-pass analysis. Should I use tools? Maybe check the directory /Users/user/jacobian3 — maybe it contains files (hints? the counterexample?). The user said "you should not look at the counterexample already found" — maybe there's a file in the directory. Let me peek at the directory listing (not reading any counterexample file contents — but even listing is fine; they said don't look at the counterexample, listing files is okay to know what's there). Actually careful: listing might reveal the answer in a filename. That's probably fine and honest to disclose. Let me just do a quick ls and mention what I found without reading solution files.

I'll keep it engaged and mathematical. Not too long though — they want to give hints next. But they also said "print as much as possible." I'll write a solid setup + landscape analysis, then ask for the hints.

Let me do the ls quickly.

Empty directory. Good — no risk of contamination. Now let me lay out my thinking.

I'll write a structured first-pass analysis:

**The problem.** F: C^n -> C^n polynomial, det JF constant nonzero (wlog = 1). Question: is F invertible with polynomial inverse? Equivalent formulations: F injective, F bijective, F proper.

**What I know that constrains where a counterexample can hide:**

1. n=1 trivial. n=2 open but Moh: deg bound (counterexample in dim 2 needs degree > 100).
2. Wang: quadratic maps are fine (deg 2 => JC true, all n).
3. Bass–Connell–Wright: reduction to cubic; Drużkowski: reduction to F = x + (Ax)^{*3} with A nilpotent-related (JA nilpotent). So a counterexample exists somewhere iff one exists in this very rigid family — but possibly in huge dimension.
4. Over R it's false without constancy (Pinchuk 1994): nonvanishing but nonconstant Jacobian, non-injective real polynomial map. Pinchuk's mechanism: the "asymptotic variety" — the set of points at infinity where F fails to be proper. Over C, det JF = const ≠ 0 forces the map to be a local biholomorphism, but it can fail to be a covering over a co-dim ≥ 2 set (the non-properness locus / Jelonek's set S_F). Key theorem (Jelonek): for a generically-finite polynomial map F: C^n -> C^n, the set S_F of points where F is not proper is either empty (then F is an automorphism, up to degree) or has pure dimension n-1 and is C-uniruled (a union of affine lines / C* -parametrized... actually (n-1)-dimensional ruled). And JC <=> S_F is empty.

So the "search space" for a counterexample: polynomial local biholomorphism whose non-proper set S_F is a nonempty hypersurface consisting of lines... but wait, det JF = 1 means the image of F is dense open and the complement of the image has codim ≥ 2? Hmm: for a local biholomorphism, F(C^n) is open; the complement of the image: a point y not in image — the fiber is empty. Known: image is dense and its complement, if nonempty... For local diffeo, I recall the image is cofinite? No — that's not right either. Let me think: Pinchuk-type phenomena over C with constant Jacobian are exactly what's unknown. There's a result that complement of the image has codimension ≥ 2 (since the map is a covering onto its image and Zariski-open image with simply-connected-ish complement argument?). Actually: a local biholomorphism F: C^n -> C^n is a covering map onto its image (this uses det bounded away... no). Hmm, there's a classical fact: an étale map F: A^n -> A^n has image containing a dense open set, and the complement of the image has codimension ≥ 2? I believe for étale F: A^n -> A^n, # (A^n \ F(A^n))... there's a theorem that an injective étale endomorphism of A^n is an automorphism (Ax-Grothendieck gives injective => bijective for varieties over C!). Ax–Grothendieck: injective polynomial self-map of C^n is bijective (but inverse may not be polynomial? For C^n injective étale, inverse is... Ax-Grothendieck + the inverse being regular: an injective regular map A^n -> A^n is an automorphism — yes, known result). So the ONLY way a counterexample can fail is non-injectivity: two distinct points mapping to the same value. That's the target: find F étale (det JF = 1) with F(p) = F(q), p ≠ q.

Equivalent: the "difference variety" — F(p) - F(q) = 0 off the diagonal has a solution. By the effective/algebraic formulation: JC true iff no such pair exists.

Also useful: Vitushkin, Orevkov tried to construct counterexamples via "etic" maps; Orevkov had a program using ball quotients... There's also the approach via "Newton polygons" in dim 2, and the crucial structural fact in dim 2 (Abhyankar-Moh / Suzuki): if F = (f,g) étale on C^2 then f has at most 2 points at infinity... and if one point at infinity then AM-suzuki says f is a variable. So a 2D counterexample must have exactly two points at infinity, degree > 100 (Moh), etc.

Also the "two-dimensional" heuristic argument FOR the conjecture: C^2 étale-cover of C^2... C^2 is not simply connected in Zariski sense... 

And crucially, the char p failure F(x) = x + x^p tells us the mechanism "derivative 1 but noninjective" exists in char p. In char 0 the analog fails because c + c^p = 0 has no nonzero root issues... In char p, x + x^p: F(a)-F(b) = (a-b) + (a-b)^p. Setting c = a-b: c^{p-1} = -1 solvable. So the map "wraps". Any char-0 counterexample must exploit a more subtle resonance.

Strategy for a real attempt: use the Drużkowski form F(x) = x + (Ax)^{*3} in dimension n with nilpotent JA, and study when F(p) = F(q): p - q = (Aq)^{*3} - (Ap)^{*3}. Let d = p - q ≠ 0. Then d = (Aq)^{*3} - (Aq + Ad)^{*3} = -3 (Aq)^{*2} (Ad) - 3 (Aq) (Ad)^2 - (Ad)^3 componentwise. Hmm — the classical approach: F injective iff the cubic map has no "difference root". There's the notion of "Meisters' problem": does JA nilpotent imply F injective for cubic Drużkowski maps? I recall it's known for n ≤ 4 or n ≤ 5? Drużkowski maps: JC verified for n ≤ 4? Hmm, I recall "Drużkowski maps in dimension ≤ 4 are injective" or similar results by Hubbers. Also "symmetric Drużkowski" (A symmetric) cases relate to... 

Actually there's a beautiful known fact: for F = x + H cubic homogeneous, injectivity can fail only if... the map F - F: consider G(x,y) = F(x) - F(y). Solutions off diagonal x ≠ y. 

OK. I shouldn't write an entire treatise before hints arrive. The user said: "Understood? I will also provide you with some hints when you are ready... once given hints devise a plan then realize it." So the right move: confirm understanding, give my honest status + a compact map of the landscape showing where a counterexample must live, propose my initial plan(s), and ask for the first hint. Print generously but don't ramble forever.

Let me also set my plan:

Plan A (algebraic search in Drużkowski family): pick small nilpotent structures, compute the difference variety, look for off-diagonal zeros, using Macaulay2/Singular/SymPy. Expectation: all small dims provably injective; the point is to build intuition for the pattern that breaks at higher n.

Plan B (non-properness/Jelonek): try to construct F étale whose S_F is a line-ruled hypersurface; study how Pinchuk's real example fails to complexify, and try to "patch" it.

Plan C (dim-2 high degree, Moh boundary): out of reach computationally, but understand the two-points-at-infinity structure.

I'll present and ask for hints. Keep formulas ASCII.

Let me parse the hints carefully.

Hint 1: coordinate degrees in output order are (7, 6, 4). So F = (F1, F2, F3): C^3 -> C^3 with deg F1 = 7, deg F2 = 6, deg F3 = 4. Max degree 7.

Wait — but I said Moh: dim 2 needs deg > 100. This is dim 3. Interesting. Hmm, wait, actually there IS a famous known counterexample-like construction in dim 3... Yes! Pinchuk's map is in R^2, non-constant Jacobian. But there's a classical construction: any Pinchuk-type map can be converted... Hmm. Actually, wait. There's a well-known fact: if you have an etale map in lower dimensions... 

Actually hold on. There's a classical construction: the Jacobian conjecture would be false if there exists a "non-injective etale map"... The hint degrees (7,6,4) in C^3... This rings a bell! I recall that from Pinchuk's real counterexample, one can construct a polynomial map in higher dimension with constant Jacobian that's not injective? No wait — Pinchuk's map has nonconstant Jacobian, so it doesn't directly apply.

Hmm, but actually — I recall there's a known example by... Let me think about "degrees (7,6,4)". There's the famous **Vénéreau polynomials** / **Vénéreau's example**: y + x(xz + y^2)... The Vénéreau fibrations: f1 = y + x^2 z + x y^2? Something like that. Vénéreau's map relates to the Dolgachev-Weisfeiler conjecture. Not it.

Degrees (7,6,4)... Also there's the famous **candidate counterexample by ... hmm. Actually! I recall: "Gorni–Zampieri" constructed examples. And there's the famous example of an etale map that is not injective in char p in dimension 3 with degrees... hmm.

Wait — actually, I strongly suspect the intended construction here is the following classical one, which turns a **real** Pinchuk map (nonvanishing Jacobian, non-injective) into a **complex** constant-Jacobian non-injective map by adding a dimension:

Given a map (p, q): C^2 -> C^2 with Jacobian determinant J(x,y) nowhere... hmm, over C a polynomial that's nowhere zero on C^2 must be constant. Over R, Pinchuk's determinant vanishes nowhere on R^2 but as a complex polynomial it has zeros off the real locus.

Trick: given f(x,y) a polynomial with no real zeros (like 1 + (xy-1)^2 x^2 ... Pinchuk's Jacobian), consider in C^3: F(x,y,z) = (p(x,y) + z*something...). There's the standard embedding trick to make the Jacobian constant:

If J(p,q) = det of the 2x2 = e(x,y), where e is nowhere-zero on the real locus but vanishes on complex points, then define F: C^3 -> C^3 by F(x,y,z) = (p(x,y) + z * a(x,y), q(x,y) + z * b(x,y), z * e(x,y)?) Hmm.

Standard trick: to "cancel" the Jacobian factor: F(x,y,z) = (p(x,y), q(x,y) + z, z*e(x,y))? Let me compute the Jacobian of F(x,y,z) = (p(x,y) + z*u, q(x,y) + z*v, z/w?)...

Actually the known construction (used to show "Jacobian Conjecture for stably..." or "Pinchuk implies a complex non-injective etale map in higher dim"? no...). Hmm, but wait: if such a construction existed turning Pinchuk into a genuine complex JC counterexample, JC would be disproved. So this can't produce a genuine counterexample... unless the premise of this exercise is that JC IS disproved by such a construction.

Hold on. Let me reconsider. There IS a classical theorem: JC is false over R even with nowhere-vanishing Jacobian (Pinchuk). But over C with constant Jacobian... open.

But actually — wait. I recall something real here: **there is NO known counterexample to JC.** So the "counterexample found by Fable" — this exercise — the degrees (7,6,4), affine-linear in z, invariant under (x,y) -> (-x,-y), constant term 1 and one other monomial (so e(x,y) = 1 + c * x^a y^b with a+b even)...

Hmm, this really smells like the **Pinchuk complexification trick**. Let me think about what construction with these degrees exists.

Pinchuk's map (1994): In its simplest form (later versions, e.g., by others), Pinchuk maps have degree (something like 10, 25)? The original Pinchuk map: p has degree 10, q degree 25? I recall deg (10, 40)? Hmm. There were improvements: a Pinchuk map of total degree 6? Not sure.

But the hint's structure: F_i(x,y,z) = A_i(x,y) z + B_i(x,y), i=1,2,3. And "a low-degree expression invariant under (x,y)->(-x,-y), constant term 1 and only one other monomial": e(x,y) = 1 + c x^a y^b, a+b even (invariant under sign flip). E.g., e = 1 + xy, or e = 1 + x^2, e = 1 + x^2 y^2, etc.

So the construction: F(x,y,z) = (A1 z + B1, A2 z + B2, A3 z + B3) with det JF = 1.

JF = 
[ A1_x z + B1_x,  A1_y z + B1_y,  A1 ]
[ A2_x z + B2_x,  A2_y z + B2_y,  A2 ]
[ A3_x z + B3_x,  A3_y z + B3_y,  A3 ]

det is a polynomial in z of degree ≤ 2. For it to equal 1 identically, the z^2, z^1 coefficients must vanish and the constant must be 1.

z^2 coefficient: det of 
[ A1_x, A1_y, A1 ]
[ A2_x, A2_y, A2 ]
[ A3_x, A3_y, A3 ] = 0 identically. This is the "Jacobian" of the map (x,y) -> (A1,A2,A3) from C^2 to C^3 — it's the condition that A1, A2, A3 are functionally dependent: the image curve is 1-dimensional... i.e., there's a polynomial relation among A1,A2,A3. Equivalently the cross product ∇A_i × ... the condition det[A_{i,x}, A_{i,y}, A_i] = 0.

This is EXACTLY the structure of Pinchuk-type constructions! I remember now: Pinchuk's construction starts with an "asymptotic" function and builds p, q so the Jacobian is e(x,y) > 0.

Hmm OK here's the thing though: the user's claim is this is a *bona fide counterexample to JC* over... wait, over which field? I assumed C. Hmm, over R maybe? "Jacobian conjecture has a counterexample" — over R with CONSTANT Jacobian? Also open? Actually the real JC with constant nonzero Jacobian: also open in general I believe (real JC). Hmm, actually the "real Jacobian conjecture" (det nowhere zero) is false by Pinchuk; with constant Jacobian over R it's still open?? I believe constant-Jacobian real JC is also open... but a complex counterexample implies a real one.

So the exercise wants: F: C^3 -> C^3 (deg 7,6,4), affine in z, det JF = 1, not injective.

But hold on — if this existed, JC would be dead and it'd be huge news. The premise says Fable found it. Since my task is to "find the logical reasoning," maybe the intended construction is real and verifiable! Let me try to actually construct it. If I can construct and verify with sympy (det JF = 1 symbolically, and exhibit p≠q with F(p)=F(q)), then either I've rediscovered a famous hidden counterexample (would be astonishing) or more likely... it doesn't exist and the premise is false. But wait — the user explicitly says "it may seem like there is no counter example possible at times but there is."

Hmm, let me think about whether a (7,6,4) construction is actually plausible, because it might secretly be a known FALSE example — i.e., the famous attempted counterexamples. Actually! I now recall: there IS a well-known **incorrect counterexample** or a known construction: "Essen–... " Hmm.

Alternatively, maybe this is the construction by **Z. Jelonek** of an etale non-injective map on a *singular* variety, or on C^3 minus something. Or the famous **"bijective but not polynomially invertible"**? No — over C bijective etale is fine.

Wait, actually, maybe I should recall: **JC in dimension 3 for maps of degree ≤ something**? There are results: JC true for n=3, deg F ≤ 2 (Wang). For cubic maps in dim 3? "n=3, deg 3" — I believe it's proven true (by various: "the 3-dimensional cubic case" maybe by ... Drużkowski/Rusek? or "Wang's theorem + ..."?). Hmm, actually I recall JC holds for n = 3 and deg F ≤ 3? Not sure. If deg ≤ 7 in dim 3 — definitely not covered by a general theorem AFAIK.

Now — the concrete construction attempt. Let me think about how to build F affine in z with det = 1.

Let me write:
F1 = A1 z + B1, F2 = A2 z + B2, F3 = A3 z + B3, all A_i, B_i in C[x,y].

det JF = sum over permutations... Let's use multilinearity in rows: each row i = z*α_i + β_i where α_i = (A_i,x, A_i,y, A_i), β_i = (B_i,x, B_i,y, B_i).

det = z^3 det(α1,α2,α3)... wait no, each ROW is split: det(z α_1 + β_1; z α_2 + β_2; z α_3 + β_3) = z^3 det(α1,α2,α3) + z^2 [det(α1,α2,β3) + det(α1,β2,α3) + det(β1,α2,α3)] + z [det(α1,β2,β3)+det(β1,α2,β3)+det(β1,β2,α3)] + det(β1,β2,β3).

For this to be ≡ 1:
(I) det(α1,α2,α3) = 0. But α_i = (A_ix, A_iy, A_i). Note det(α1,α2,α3) = sum over i,j cyclic: A_1 (A_2x A_3y - A_2y A_3x) - ... = A1·J(A2,A3) - A2·J(A1,A3) + A3·J(A1,A2), where J(f,g) = f_x g_y - f_y g_x.

Hmm wait, more carefully: det of matrix with rows (A_ix, A_iy, A_i):
= A_1x (A_2y A_3 - A_3y A_2) - A_1y (A_2x A_3 - A_3x A_2) + A_1 (A_2x A_3y - A_2y A_3x)
= -A_3 [A_1x A_2y - A_1y A_2x] ... let me just collect: coefficient of A_3: A_1x A_2y... hmm from first term: A_1x A_2y A_3... wait first term has A_2y A_3 and -A_3y A_2; the -A_1y(...A_2x A_3...) gives -A_1y A_2x A_3. And A_1(A_2x A_3y - A_2y A_3x). So total = A_3 (A_1x A_2y - A_1y A_2x) - A_2 (A_1x A_3y - A_1y A_3x) + A_1 (A_2x A_3y - A_2y A_3x) = A_3 J(A1,A2) - A_2 J(A1,A3) + A_1 J(A2,A3). Yes.

Condition (I): A1 J(A2,A3) + A2 J(A3,A1) + A3 J(A1,A2) = 0.

This is the condition that the map (x,y) -> (A1,A2,A3) has image of dimension ≤ 1... Actually it's the condition that the vector field... It's the Wronskian-like condition: the three functions A_i are functionally dependent along with... Hmm: for a map A: C^2 -> C^3, image is a curve iff any two components are dependent, i.e., J(A_i, A_j) = 0 for all pairs. But here the condition is weaker: A · (J(A2,A3), J(A3,A1), J(A1,A2)) = 0, i.e., A is orthogonal to the "cross product" ∇A_i... The vector v = (J(A2,A3), J(A3,A1), J(A1,A2)) is normal to the image surface; condition (I) says A(x,y) lies in the tangent plane of the image at A(x,y). Interesting geometric condition!

Alternative: think of it as: the polynomial map G(x,y,z) = z*(A1,A2,A3) : C^3 -> C^3 maps into... the cone over the image curve/surface, and condition (I) says the image of G is contained in a 2D variety (the tangent condition...). Hmm.

(IV) det(β1, β2, β3) = det J(B1,B2,B3) with third column B_i... wait β_i = (B_ix, B_iy, B_i)?? No wait. Hold on: row i of JF is (∂F_i/∂x, ∂F_i/∂y, ∂F_i/∂z) = (A_ix z + B_ix, A_iy z + B_iy, A_i). So α_i = (A_ix, A_iy, A_i) and β_i = (B_ix, B_iy, 0). 

Ah, that simplifies things a lot! β_i = (B_ix, B_iy, 0). Then:

det(β1,β2,β3) = 0 automatically (third column zero in all rows → determinant zero). Good — so the z^0 term is 0?? But we need det JF = 1! So the constant term must come from... wait no. det JF = z^3 T3 + z^2 T2 + z T1 + T0 where T0 = det(β1,β2,β3) = 0. That can't be right — then det JF would vanish at z=0, impossible for = 1.

Hmm wait, I mislabeled. The determinant is linear in each row: row i = z α_i + β_i. The expansion is over choosing for each row either z α_i or β_i. Terms:
- z^3: det(α1, α2, α3)
- z^2: det(α1,α2,β3) + det(α1,β2,α3) + det(β1,α2,α3)
- z^1: det(α1,β2,β3) + det(β1,α2,β3) + det(β1,β2,α3)
- z^0: det(β1,β2,β3)

With β_i = (B_ix, B_iy, 0): det(β1,β2,β3) = 0 (last column all zeros). And z^1 terms: e.g. det(β1, β2, α3) = det of [[B1x, B1y, 0],[B2x, B2y, 0],[A3x, A3y, A3]] = A3 (B1x B2y - B1y B2x) = A3 J(B1,B2). Similarly det(β1, α2, β3) = A2 J(B1, B3)·(sign?) Let me compute: rows (B1x,B1y,0), (A2x,A2y,A2), (B3x,B3y,0). Expanding along third column: only the middle row contributes: -A2 · det[[B1x,B1y],[B3x,B3y]] · (sign for position (2,3)) = -A2 (B1x B3y - B1y B3x) = -A2 J(B1,B3). Hmm sign: cofactor C_{23} = (-1)^{2+3} M_{23} = -M_{23}. det = sum a_{i3} C_{i3} = a_{23} C_{23} = A2 · (-(B1x B3y - B1y B3x)) = -A2 J(B1,B3). And det(α1,β2,β3): third column only row 1: position (1,3), sign (+): A1 (B2x B3y - B2y B3x) = A1 J(B2,B3).

So z^1 coefficient: T1 = A1 J(B2,B3) - A2 J(B1,B3) + A3 J(B1,B2) = A1 J(B2,B3) + A2 J(B3,B1) + A3 J(B1,B2).

Similarly z^2: T2 = det(α1,α2,β3) + det(α1,β2,α3) + det(β1,α2,α3). Third column of β3 is 0, so det(α1,α2,β3) = 0. Similarly det(α1,β2,α3) = 0 (β2 has 0 in third position). And det(β1,α2,α3) = 0 (β1 third entry 0). So T2 = 0 automatically!

And T3 = det(α1,α2,α3) = A1 J(A2,A3) + A2 J(A3,A1) + A3 J(A1,A2) [same formula as before].

So the full condition det JF ≡ 1:

(a) T3 = A1 J(A2,A3) + A2 J(A3,A1) + A3 J(A1,A2) ≡ 0.
(b) T1 = A1 J(B2,B3) + A2 J(B3,B1) + A3 J(B1,B2) ≡ 1.

Beautiful — this is a clean system! And note the antisymmetric structure. 

Now, condition (a): the tangent-plane condition for the map A = (A1,A2,A3): C^2 -> C^3. Simplest way to satisfy: make the image of A a curve, i.e., all J(A_i, A_j) = 0 (A_i functionally dependent). E.g., A2, A3 functions of A1. Then (a) holds trivially.

Condition (b) then needs A1 J(B2,B3) + A2 J(B3,B1) + A3 J(B1,B2) = 1.

Simplest instance: choose A3 = 0? Then F3 = B3(x,y) — doesn't involve z. Hmm, but then for F to have any chance... F3 = B3(x,y) with deg 4 per the hint. And (b): A1 J(B2,B3) + A2 J(B3,B1) = 1.

Hmm wait, but actually with A3 = 0, condition (a): A1 J(A2, 0) + ... = 0 trivially regardless. And T1 = A1 J(B2,B3) + A2 J(B3,B1) + 0 = 1.

Hmm, but is A3 = 0 wise? Then z only enters F1, F2. det JF = ... let's sanity check directly: F3 = B3(x,y). JF = [A1x z + B1x, A1y z + B1y, A1; A2x z + B2x, ..., A2; B3x, B3y, 0]. det = expand along third row: B3x · (A1y z + B1y)A2 ... = B3x·(A2 (A1y z + B1y) - A1(A2y z + B2y))·(+? position (3,1) sign +) ... = B3x [z(A1y A2 - A1 A2y) + (A2 B1y - A1 B2y)] - B3y [z(A1x A2 - A1 A2x) + (A2 B1x - A1 B2x)] = z[A1 J(A2, B3)... let me just: z·[A2(A1y B3x - A1x B3y) - A1(A2y B3x - A2x B3y)] + [A2(B1y B3x - B1x B3y) - A1(B2y B3x - B2x B3y)] = z·[A1 J(A2,B3) + A2 J(B3,A1)] + [A1 J(B2,B3) + A2 J(B3,B1)]. 

Setting z-coefficient 0: A1 J(A2,B3) = A2 J(A1,B3)... i.e., A1 J(A2, B3) - A2 J(A1, B3) = 0, i.e., J(A1/A2? ...). This says J(A2, B3)/A2 = J(A1, B3)/A1 — condition that B3 is "equally dependent" on A1 and A2. Simplest: A1, A2, B3 all functions of one variable t(x,y). Then constant part: A1 J(B2,B3) + A2 J(B3,B1) = 1.

OK regardless of these simplifications, the hint says: use a low-degree expression e(x,y) = 1 + (one monomial), invariant under (x,y)->(-x,-y), i.e., e = 1 + c x^a y^b, a+b even.

The classic Pinchuk construction (as simplified by various authors; I recall a version due to Pinchuk with deg (10,25), then improvements, e.g., by "van den Essen" in his book, there's a Pinchuk example with degrees... In van den Essen's book "Polynomial Automorphisms and the Jacobian Conjecture", the Pinchuk example given has deg p = 10, deg q = 25, and Jacobian = something like... hmm. There's also a degree (6, ?) version? And there's the famous observation that no Pinchuk map can have constant positive Jacobian? No wait — Pinchuk maps have Jacobian a polynomial positive on R^2 but nonconstant.

Hmm, but here we want CONSTANT Jacobian in dimension THREE, and the degrees (7,6,4). 

OH WAIT. I think I remember now! There's a classical construction: **from a Pinchuk map (p,q) with Jacobian e(x,y) (nonvanishing on R^2, even-ish structure), build a complex map in one dimension higher with constant Jacobian that is not injective** — and this DOESN'T disprove JC because... hmm, no, it would disprove JC if it worked. Unless the resulting map's non-injectivity... no.

Hmm hold on, actually, let me reconsider: maybe I DO recall a counterexample-shaped object: the **"Gorni-Zampieri / Pinchuk" example over R^3?** Or... Actually! I remember: there's a famous example of a **local diffeomorphism F: R^3 -> R^3 with det JF = 1 everywhere that is not injective**? Hmm, no — wait, for CONSTANT Jacobian over R, injectivity... there might be counterexamples in the real setting with constant Jacobian?! Actually YES — I recall: "The real Jacobian conjecture is false even with constant Jacobian in dimension ≥ 3"? Hmm, is that a thing? Pinchuk's is dim 2 nonconstant. Hmm, I genuinely am not sure.

Alternatively — the famous **"Feller–... "** no. Let me think about **Vitushkin's example**: Vitushkin constructed etale maps between affine varieties... Orevkov: "An example in connection with the Jacobian conjecture" (Orevkov constructed a non-injective etale map from a certain smooth affine threefold (a Z/5 quotient thing) to C^3?). Yes! Orevkov's example: an étale, non-injective map from a smooth affine variety X (homology of C^3, a Z/5-homology ball thing) onto C^3... that's the "fake C^3" direction (Ramanujam, tom Dieck–Petrie). Not a counterexample to JC itself.

Hmm hmm. But actually, the DEGREES (7,6,4) really ring a bell with Pinchuk-style maps in dimension 3. Let me just try to construct it. Actually, you know what — let me reconsider. I recall a paper: **"A counterexample to the strong real Jacobian conjecture" (Pinchuk)** and separately **F. Braun, J.R. dos Santos Filho** "The real jacobian conjecture on R^2 is true when one of the components has degree 3". And also: **"Pinchuk maps" have the form**: with auxiliary polynomials t = xy - 1, h = t(xt+1), f = (xt+1)^2 (t^2 + y)... something like that. Let me recall Pinchuk's original construction:

Pinchuk 1994: Let t = xy - 1, h = t(xt + 1), f = (xt + 1)^2 (t^2 + y), and p = h + f, q = -t^2 - 6th(h+1) - 170 fh - 91 h^2 - 195 f h^2 - 69 h^3 - 75 f h^3 - (75/4) h^4. Then J(p,q) = ... equals some polynomial positive on R^2 (sum of squares: I recall J = t^2 + (something)^2 + ...). Degrees: deg p = 10, deg q = 25? The Jacobian: I recall J(p,q) = t^2 + (t(t+h) + f)^2·(...) + ... something manifestly positive on R^2.

The key structural feature: t = xy - 1 is invariant under (x,y) -> (-x,-y)! Because (-x)(-y) = xy. YES. And "constant term 1"? t = xy - 1 has constant term -1... but e = 1 - xy or e = 1 + xy also invariant. The hint says: "a low-degree expression invariant under (x,y) -> (-x,-y), constant term 1 and contains only one other monomial" → e(x,y) = 1 + c x^a y^b with a+b even — like 1 - xy = -t. Indeed -t = 1 - xy. 

So the construction is DEFINITELY Pinchuk-flavored: build everything from e = 1 - xy (or 1 + xy) — invariant under the sign flip, so F(-x,-y,z) relates to F(x,y,z) — and non-injectivity comes precisely from that symmetry: F(p) = F(-p)? No wait, if all components were invariant under (x,y)->(-x,-y), then F(x,y,z) = F(-x,-y,z') might give collisions. Hmm, but e invariant alone doesn't make components invariant.

Let me now think about what the actual target is. I'll now seriously consider that this "counterexample" is a REAL published (but later found erroneous?) example OR a genuine counterexample to a real-JC variant. Degrees (7,6,4), dim 3, affine in z, constant Jacobian 1, non-injective over... let me just try to construct one and TEST it. If I succeed, great — verify computationally. If it's real, it would disprove JC... The user insists there is one. My job: find the reasoning, construct, verify. Let me actually attempt the construction seriously, because the hint structure gives a concrete system of equations (a), (b) above, plus non-injectivity via the (x,y)->(-x,-y) symmetry.

Plan: 
1. Set up the ansatz with e = 1 - xy (or 1 + xy).
2. Use the z-affine structure with A3 = 0? But hint says F3 has degree 4 and is affine-linear in z — maybe F3 = A3 z + B3 with A3 =/= 0. Hmm, degrees: F1 deg 7, F2 deg 6, F3 deg 4.
3. Non-injectivity: want points p =/= q with F(p) = F(q). Symmetry suggests q = (-x, -y, z') for some z'.

Let me think: if B_i(x,y) are all ODD functions under (x,y)->(-x,-y) and A_i are EVEN, then F_i(-x,-y,z) = A_i(x,y) z - B_i(x,y). So F(-x,-y,z) = (A z - B) and F(x,y,-z) = -A z + B = -(A z - B) = -F(-x,-y,z). Interesting: F(x,y,-z) = -F(-x,-y,z) if A even, B odd. Then F(-x,-y,z) = -F(x,y,-z). If we could find (x,y,z) with F(x,y,z) = -F(x,y,-z)... i.e. B(x,y) = 0 and ... hmm: F(x,y,z) = -F(x,y,-z) iff A z + B = -(-A z + B) = A z - B iff B = 0. So on the locus B1 = B2 = B3 = 0 (some curve in (x,y)), F(x,y,z) = z(A1,A2,A3) and F(x,y,-z) = -z(A1,A2,A3). Then F(x,y,z) = F(-x,-y,-z)? F(-x,-y,-z) = A(-x,-y)(-z) + B(-x,-y) = -z A(x,y) - B(x,y) = -(zA + B) = -F(x,y,z). So F(-p) = -F(p) for all p. Odd map. For a collision we'd want F(p) = F(p') with p' =/= ±p.

Alternatively, maybe non-injectivity: find (x,y) with A(x,y) = 0 vector? Hmm.

Let me think again about Pinchuk's mechanism: the non-injectivity in Pinchuk's map comes from two points, often (x,y) and (-x,-y)-ish... Actually in Pinchuk's example, the two colliding points are found where the "fiber curve" wraps around.

Honestly, the most productive path: actually DO the algebra/computation. Let me set up the system in sympy and try to solve a concrete ansatz. This is what the user wants — print reasoning, then compute.

Let me design the ansatz more concretely using the degrees (7,6,4):

Constraints:
- F1 = A1 z + B1, deg 7.
- F2 = A2 z + B2, deg 6.
- F3 = A3 z + B3, deg 4.
- (a) A1 J(A2,A3) + A2 J(A3,A1) + A3 J(A1,A2) = 0.
- (b) A1 J(B2,B3) + A2 J(B3,B1) + A3 J(B1,B2) = 1.

Simplest way to kill (a): make A1, A2, A3 pairwise dependent: all functions of a single polynomial t(x,y). Say A_i = a_i(t) (polynomials in one variable t). Then all J(A_i, A_j) = 0, so (a) holds automatically!

Then (b): A1 J(B2,B3) + A2 J(B3,B1) + A3 J(B1,B2) = 1. Note J(B_i, B_j) = B_i,x B_j,y - B_i,y B_j,x.

Suppose further that B_i = b_i(t, s) where s is another polynomial, or even simpler: suppose the B's are built so that the J-terms produce constants times powers of t... The right side must be the constant 1.

Now bring in e = 1 - xy. Take t = xy - 1 (so e = -t) or t = xy. Hmm, hint: e has constant term 1 and one other monomial: e = 1 + c x^a y^b, a+b even. Simplest: e = 1 + xy? or 1 - xy, or 1 + x^2, 1 + x^2 y^2, 1 + x^4...

In Pinchuk's original: t = xy - 1, i.e., -e with e = 1 - xy. Everything (h, f, p, q) is built from t, and the "parity" structure: xt + 1 = x(xy-1) + 1 = x^2 y - x + 1 — this is "even" in the sense xt+1 is invariant under (x,y)->(-x,-y): (-x)((-x)(-y)-1)+1 = -x(xy-1)+1 = -(xt) + 1... no: (-x)(t) + 1 where t same = -xt + 1. Not invariant. Hmm. But x^2 y stuff...

Let me look at degrees to guess the shape. If t = xy (deg 2):
- A_i = polys in t: to get deg(A1 z + B1) = 7: deg A1 ≤ 7 (as poly in x,y; A1 = a1(t) with deg a1 = d gives deg 2d; so A1 of degree 6 means a1 cubic in t; but 7 is odd — odd total degree requires B1 to have odd-degree monomials, like x·(even stuff).

Degree 7 with even/odd mix: B1 could contain x * (deg 6 even part). E.g., B1 = x * c(t) where c(t) has deg ≤ 3 in t (deg 6 in x,y). Similarly F2 deg 6: could be pure even (poly in t of deg 3, or x^2·stuff). F3 deg 4: even, poly in t deg 2, or y·(deg 3 odd)... deg 4 even: could be poly in t of degree 2 (deg 4) — like h = t(xt+1)? deg 4. YES: h = t(xt+1) = x t^2 + t = x(xy)^2 + xy = x^3 y^2 + xy: degree 4! That matches F3 deg 4 maybe? And in Pinchuk, h = t(xt+1) has degree... t deg 2, xt+1 deg 3 → h deg 5? Wait: xt+1 = x(xy-1)+1 = x^2y - x + 1: degree 3. h = t(xt+1): degree 2+3 = 5? Hmm t(xt+1) = x t^2 + t: x t^2 has degree 1 + 4 = 5. So h deg 5. Not 4.

Let me not fixate on reconstructing Pinchuk's exact map. Instead, let me think about what makes the Jacobian constant and non-injectivity work, guided by the symmetry.

KEY INSIGHT about non-injectivity in Pinchuk-type constructions: The map is built from functions of t = xy-1 (even under flip) and x (odd). Under (x,y) -> (-x,-y): t fixed, x flips. So a component like p = h + f where... In Pinchuk's map, the collision happens at two points related by this flip with different z-values compensating.

For our 3D z-affine map: F_i(x,y,z) = A_i(t,...) z + B_i. If A_i are functions of t only (even) and B_i odd under flip, then F(-x,-y,z) = A z - B, and F(x,y,-z) = -Az + B = -(Az - B). So F(x,y,-z) = -F(-x,-y,z). A collision F(x,y,z) = F(x',y',z') could be sought with (x',y',z') = (-x,-y,w): need A z + B = A w - B, i.e., 2B = A(w - z). Three equations: 2 B_i = A_i (w - z) for i = 1,2,3. So if at some (x,y) the vector B(x,y) is parallel to A(x,y) — i.e., B_i / A_i all equal to some λ — then w = z + 2λ gives a collision F(x,y,z) = F(-x,-y,w) with (x,y,z) =/= (-x,-y,w) as long as (x,y) =/= (0,0)-fixed points of flip... the flip has no fixed points except x=y=0. So we need a point (x,y) where B = λ A as vectors in C^3 and A =/= 0.

That's a very clean mechanism!! Non-injectivity <=> exists (x,y) in C^2, λ in C: B_i(x,y) = λ A_i(x,y) for all i (with A =/= 0 there; if A = 0 and B = 0 then F(x,y,z) constant in z... wait if A(x,y)=0 as a vector and B(x,y) = 0 vector, then F(x,y,z) = 0 for all z — infinitely many preimages of... (0,0,0)? F(x,y,z) = (0,0,0) for all z — that's already non-injective, even stronger).

Hmm wait, but actually even B = λA with A=0, B=0... Let me reconsider: collision condition F(x,y,z) = F(-x,-y,w) with B odd, A even:
F_i(x,y,z) = A_i z + B_i; F_i(-x,-y,w) = A_i w - B_i. Equality: A_i(z - w) = -2B_i, i.e., A_i (z-w) + 2B_i = 0 for all i. If we pick (x,y) and λ := (w - z)/2 such that B_i(x,y) = λ A_i(x,y) ∀i, collision. Note we need the SAME λ for all three coordinates — B(x,y) parallel to A(x,y) in C^3. The locus "B parallel to A" is given by the 2×2 minors: B1 A2 - B2 A1 = 0, B1 A3 - B3 A1 = 0 — two equations in (x,y) in C^2: expected 0-dimensional, i.e., solutions EXIST generically!! 

This is the heart: over C, two equations in two variables generically have solutions (unless they conspire to be inconsistent — e.g., resultant a nonzero constant). The whole art of the construction: arrange det JF = 1 (conditions (a),(b)) while NOT conspiring away the parallel-locus solutions. And nilpotency/etale-ness constrains things heavily — but apparently the (7,6,4) example threads the needle.

Hmm wait, but hold on: if such an example genuinely existed, JC would be FALSE. This is exactly why I'm suspicious — but the exercise says go find it. And the algebra above shows the counterexample conditions are remarkably mild: ansatz A_i = a_i(t) (kills (a) automatically), B odd under flip, and condition (b): A1 J(B2,B3) + A2 J(B3,B1) + A3 J(B1,B2) = 1. Then find the parallel locus.

Now: is condition (b) achievable with deg constraints (A1 z + B1 deg 7, etc.)? Let me think about degrees. Let t have degree 2 (t = xy or xy - 1; e = 1 - xy or 1 + xy).

For (b) to be constant = 1 with A_i = a_i(t): J(B_i, B_j) should give things like 1/a_k(t)-ish... degrees must cancel. Let's think: we need sum of three terms to be a nonzero constant. Since A_i are polys in t of positive degree (probably), we need J(B_j, B_k) to involve NEGATIVE... no, polynomials only. So cancellation among the three terms must reduce to constant. E.g., suppose B_i = b_i(t) u + c_i(t) where u is odd (like x or y) ... let's compute J(B_i, B_j) for B_i = P_i(t) x (odd under flip, since x odd, t even):

B_i = P_i(t) x, t = xy (say). B_i,x = P_i'(t) y x + P_i(t); B_i,y = P_i'(t) x^2.
J(B_i, B_j) = (P_i' y x + P_i)(P_j' x^2) - (P_i' x^2)(P_j' y x + P_j) = x^2 [P_i P_j' - P_i' P_j] = x^2 · W(P_i, P_j)(t) where W is the Wronskian in t.

Then condition (b) = A1 x^2 W(P2,P3) + A2 x^2 W(P3,P1) + A3 x^2 W(P1,P2) = x^2 · [sum a_i(t) W(...)(t)] = 1?? LHS divisible by x^2 — can't be 1. Unless B's aren't all proportional to x. So pure "P(t) x" ansatz fails. Need a mix like B_i = P_i(t) x + Q_i(t) y? Let's compute: B_i = P_i x + Q_i y with t = xy.

B_i,x = P_i' y x + P_i + Q_i' y^2; B_i,y = P_i' x^2 + Q_i' x y + Q_i.
J(B_i,B_j) = (P_i' t + P_i + Q_i' y^2)(P_j' x^2 + Q_j' t + Q_j) - (P_j' t + P_j + Q_j' y^2)(P_i' x^2 + Q_i' t + Q_i).

This is getting messy but symmetric. The x^2 y^2 = t^2 terms: from (P_i' t)(Q_j' t)... wait P_i' t · Q_j' ... no: first factor's (P_i' t) times second factor's (Q_j' t)? Second factor terms are in x^2, t, and constant-ish... hold on, I mis-set: B_i,y = P_i' x^2 + Q_i' x y + Q_i where xy = t. So B_i,y = P_i' x^2 + Q_i' t + Q_i.

J = B_i,x B_j,y - B_j,x B_i,y. The y^2 · x^2 = t^2 cross terms: (Q_i' y^2)(P_j' x^2) - (Q_j' y^2)(P_i' x^2) = t^2 (Q_i' P_j' - Q_j' P_i') = t^2 W(Q,P)... wait = t^2 (Q_i' P_j' - P_i' Q_j'). The (P_i' t + P_i)(Q_j' t + Q_j) - (P_j' t + P_j)(Q_i' t + Q_i) terms: expand: P_i' Q_j' t^2 + P_i' Q_j t + P_i Q_j' t + P_i Q_j - [swap] = t(P_i' Q_j - P_j' Q_i) + t (P_i Q_j' - P_j Q_i') + (P_i Q_j - P_j Q_i). The t^2 cancels between these two. So total:

J(B_i,B_j) = t^2 (Q_i' P_j' - P_i' Q_j') + t (P_i' Q_j - P_j' Q_i + P_i Q_j' - P_j Q_i') + (P_i Q_j - P_j Q_i) + cross terms with x^2: (P_i' t + P_i)(P_j' x^2) - (P_j' t + P_j)(P_i' x^2) = x^2 (P_i P_j' - P_i' P_j). Wait I forgot: first factor B_i,x has (P_i' t + P_i) + Q_i' y^2; B_j,y = P_j' x^2 + (Q_j' t + Q_j). So cross terms: (P_i' t + P_i)(P_j' x^2) + (Q_i' y^2)(Q_j' t + Q_j) - swap. The second pair: y^2 (Q_i' (Q_j' t + Q_j) - Q_j' (Q_i' t + Q_i)) = y^2 (Q_i' Q_j - Q_j' Q_i) = -y^2 W(Q_i, Q_j)... = y^2 (Q_i' Q_j - Q_i Q_j').

So J(B_i, B_j) = x^2 W(P_i, P_j) + t^2 W(Q_i, P_j)... let me recompute signs: W(f,g) := f g' - f' g.
- x^2 term: P_i P_j' - P_i' P_j = W(P_i, P_j). ✓.
- y^2 term: Q_i' Q_j - Q_j' Q_i = -W(Q_i, Q_j).
- t^2 term: Q_i' P_j' - P_i' Q_j'... from (Q_i' y^2)(P_j' x^2) - (Q_j' y^2)(P_i' x^2) = y^2 x^2 (Q_i' P_j' - Q_j' P_i') = t^2 (Q_i' P_j' - Q_j' P_i'). Hmm = t^2 · (Q_i' P_j' - P_i' Q_j').
- t terms: t (P_i' Q_j - P_j' Q_i) + t (P_i Q_j' - P_j Q_i') = t [W(Q_j, P_i) + W(P_i, Q_j)] = 0!! Wait: P_i' Q_j - P_j' Q_i = W(Q_j, P_i)... W(f,g) = f g' - f' g; W(Q_j, P_i) = Q_j P_i' - Q_j' P_i. Yes = P_i' Q_j - Q_j' P_i. And second: P_i Q_j' - P_j Q_i' = W(P_i, Q_j) = -W(Q_j, P_i). So the t terms CANCEL. 
- constant (in x,y) term: P_i Q_j - P_j Q_i =: D(P,Q).

So: J(B_i, B_j) = x^2 W(P_i,P_j) - y^2 W(Q_i,Q_j) + t^2 (Q_i' P_j' - P_i' Q_j') + (P_i Q_j - P_j Q_i).

Hmm wait, but I think I mislabeled—let me double check the t^2 term exists... (Q_i' y^2)·(P_j' x^2): B_i,x contains Q_i' y^2, B_j,y contains P_j' x^2. Yes. So t^2 term present.

This is getting complicated. The constant-Jacobian condition (b) with A_i = a_i(t):

sum_i a_i(t) J(B_j, B_k) = sum a_i [x^2 W(P_j,P_k) - y^2 W(Q_j,Q_k) + t^2(...) + D(P_j,P_k)] = 1.

For this to collapse to 1, need: x^2-coefficient sum = 0, y^2-coefficient sum = 0, t^2-coefficient sum = 0, and the D-sum = 1. But D(P_j,P_k) = P_j Q_k - P_k Q_j are functions of t, and sum a_i(t) D(P_j,P_k)(t) = 1 — a function of t equal to 1: possible if it's constant!

So conditions:
(Σx) sum a_i W(P_j, P_k) = 0
(Σy) sum a_i W(Q_j, Q_k) = 0
(Σt) sum a_i (Q_j' P_k' - P_j' Q_k') = 0
(Σ1) sum a_i (P_j Q_k - P_k Q_j) = 1.

Interesting. These are conditions on one-variable polynomials a_i, P_i, Q_i in t. The structure resembles a 3×3 determinant identity: for any 3 vectors... Actually! There's a slick way: consider the 3×3 matrices M = [a_i; ...]. Note sum_i a_i (P_j Q_k - P_k Q_j) cyclic = det of matrix with rows (a1,a2,a3), (P1,P2,P3), (Q1,Q2,Q3). Yes! (Σ1) = det[a; P; Q] (rows a, P, Q). Similarly (Σx) = det[a; P; P'] and (Σy) = det[a; Q; Q'], (Σt) = det[a; Q'; P'].

So the conditions are:
det[a; P; Q] = 1, det[a; P; P'] = 0, det[a; Q; Q'] = 0, det[a; Q'; P'] = 0, as polynomials in t, where a, P, Q are triples of polys in t.

det[a; P; P'] = 0 says a is in the span of P, P' over... it's the condition that the curve t -> (a(t), P(t))-ish... If a = P componentwise proportional? Simplest: choose P = a. Then det[a; P; P'] = det[P; P; P'] = 0 ✓. Similarly Q' proportional to a? If Q' = a, then det[a; Q; Q'] = det[a;Q;a] = 0 ✓ and det[a; Q'; P'] = det[a; a; P'] = 0 ✓. And (Σ1): det[a; P; Q] = det[P; P; Q]?? No wait if a = P then det[a;P;Q] = 0. Contradiction with =1. 

Try: P = a componentwise? kills (Σ1). Hmm. Need det[a;P;Q] = 1 =/= 0, so a, P, Q linearly independent as vectors of functions. But the other three dets must vanish. det[a;P;P']=0 means a, P, P' dependent; since a, P independent (else det[a;P;Q] might still be nonzero if... if a = cP then det[a;P;Q] = c det[P;P;Q] = 0 — bad). So a, P independent, and P' in span(a, P). Similarly Q' in span(a, Q). And Q', P' dependent with a: det[a;Q';P'] = 0 — follows from Q' in span(a,Q), P' in span(a,P)? Not automatically, need to check.

Simplest: P' in span over CONSTANTS of (a, P), i.e., P' = c1 a + c2 P with c_i constants (not functions). Solutions: P satisfies a linear ODE. Similarly Q' = d1 a + d2 Q.

Try simplest: c2 = d2 = 0: P' = c1 a, Q' = d1 a. Then check det[a; Q'; P'] = det[a; d1 a; c1 a] = 0 ✓. And (Σ1): det[a; P; Q] = 1. With P = c1 ∫a, Q = d1 ∫a — then P = (c1/d1) Q + const·... P and Q both integrals of a: P = c1 I + p0, Q = d1 I + q0 where I = ∫ a. det[a; P; Q] = det[a; c1 I + p0; d1 I + q0] = det[a; c1 I; q0] + det[a; p0; d1 I] + det[a; p0; q0] (using multilinearity, det[a;I;I]=0) = c1 q0 det[a; I; 1] + d1 p0 det[a; 1; I] + p0 q0 det[a;1;1] = (c1 q0 - d1 p0) det[a; I; 1] = (c1 q0 - d1 p0)(a·(I·1' ... compute det[a;I;1] = a(I·0 - 1·1) - I(...)... rows (a1,a2,a3), (I1,I2,I3), (1,1,1): det = sum a_i (I_j - I_k)... = a1(I2 - I3) + a2(I3 - I1) + a3(I1 - I2). Since I_i = ∫ a_i, this = a1 ∫(a2-a3) + ... hmm circular. This can be a nonzero function of t but making it EXACTLY 1 is restrictive. Note derivative: d/dt det[a;I;1] = det[a';I;1] + det[a; a; 1] = det[a';I;1]. For this to be 0 (det const), need det[a'; I; 1] = 0.

This is getting deep. But you know, maybe I'm overcomplicating: the B_i might not be of the form P(t) x + Q(t) y. Let me step back and think about the ACTUAL known Pinchuk-type construction in the literature with degrees (7,6,4) in dimension 3. 

AH WAIT. Now I really do remember! There's a famous example: **"the map of Gorni–Zampieri"**? Or... **"Essen's example"**? Hmm, degrees (7,6,4)... dimension 3, constant Jacobian, non-injective, built from Pinchuk's... 

Actually — I recall a specific paper: **S. Pinchuk, "On the Jacobian conjecture"** or a note where from his R^2 example he builds a C^3 (or R^3) map with constant nonzero Jacobian determinant that is not injective?? Hmm, if Pinchuk had that, JC would be disproved. So no.

BUT WAIT — maybe over R! "Real Jacobian Conjecture" with constant Jacobian in dimension 3 — hmm, is there a counterexample? I recall: **the real Jacobian conjecture with CONSTANT Jacobian is open even in dimension 2**... and I don't recall a dim-3 counterexample.

Hmm, alternatively, there's a classical construction: if (p,q): R^2 -> R^2 has Jacobian e(x,y) > 0 everywhere (Pinchuk), then F(x,y,z) = (p(x,y) + z·e(x,y)·?, ...). Hmm: Consider F(x,y,z) = (p(x,y), q(x,y) + z·e(x,y)... let me think: we want to use the extra dimension to "absorb" the nonconstant Jacobian. 

Try: F(x,y,z) = (p(x,y) + z·r(x,y), q(x,y) + z·s(x,y), z·e(x,y))? Then det JF: third row (z e_x, z e_y, e). Hmm, expand: det = e·J(p,q) - ... let's see with F3 = z e: det JF = det[p_x + z r_x, p_y + z r_y, r; q_x + z s_x, q_y + z s_y, s; z e_x, z e_y, e]. At z = 0: det = det[p_x, p_y, r; q_x, q_y, s; 0, 0, e] = e(p_x q_y s... = e(p_x q_y - p_y q_x)·... expand along third row: e(p_x q_y - p_y q_x) = e·J(p,q) = e·e = e^2. Not 1. Try F3 = z/e? Not polynomial.

Try F = (p + z u, q + z v, z/w)? nah.

Standard trick: embed into higher dim to kill the Jacobian: F(x,y,z) = (p(x,y), q(x,y) + z, z e(x,y))? det: rows (p_x, p_y, 0), (q_x + z·0... F2 = q + z: row (q_x, q_y, 1); F3 = z e: row (z e_x, z e_y, e). det = p_x (q_y e - z e_y) - p_y (q_x e - z e_x) + 0 = e J(p,q) - z (p_x e_y - p_y e_x) = e^2 - z J(p, e). Still not constant. 

Take F = (p + z, q, z e): rows (p_x, p_y, 1), (q_x, q_y, 0), (z e_x, z e_y, e). det = -[p_x (q_y e - 0) - p_y (q_x e - 0)]·... expand along 2nd row? det = -q_x(p_y e - z e_y) + q_y(p_x e - z e_x) = -e J(q,p)·... = -e(p_y q_x - p_x q_y) - z(q_x ... hmm = e J(p,q) - z(q_x p_y ... let me: = -q_x p_y e + q_x z e_y + q_y p_x e - q_y z e_x = e J(p,q) + z(q_x e_y - q_y e_x) = e^2 + z J(e, q). Nonconstant.

The clean trick is: F(x,y,z) = (p(x,y) + z e(x,y)·c1, q(x,y) + z e c2, z·? )... Hmm, alternatively use F3 = z/e requires rational.

OK here's the classical "Pinchuk ⇒ complex counterexample to JC in dim 3" FALSE-hope construction that people try: it always fails because e(x,y) has complex zeros; at a complex zero of e, the extended map's Jacobian degenerates. So the 3D map can't be Keller. UNLESS the construction is cleverer.

Given the hint structure (affine in z, degrees 7,6,4, e = 1 + monomial, sign-flip symmetry), let me now hypothesize the actual example is the following REAL, PUBLISHED object: I'm now fairly confident this is from a specific paper I should reconstruct: **"A counterexample to a conjecture on the..."** hmm. OR it's the example from van den Essen's book / Drużkowski's survey showing a Pinchuk map of low degree.

Actually — hold on, degrees! Let me recall the LOWEST-degree known Pinchuk map: I believe there's a Pinchuk map with deg(p) = 6, deg(q) = 9? or total degree 6? Hmm, "Pinchuk maps of total degree 6" — hmm. There was a paper showing Pinchuk maps must have deg ≥ 6 and constructing one of degree (6, ...)? 

And then there's the **"Gwoździewicz"** example: a Pinchuk map with p of degree 6? Hmm. Or **Campbell**'s work on Pinchuk maps: "The asymptotic variety of a Pinchuk map" and "Rational Pinchuk maps". Campbell studied the map with p = h + f etc.

Let me try yet another angle: maybe the example is genuinely a COUNTEREXAMPLE TO JC OVER A NON-ALGEBRAICALLY-CLOSED FIELD or in POSITIVE CHARACTERISTIC, or over R. The user said "Jacobian conjecture ... counterexample". The most famous true theorem of this shape: **the real-analog is false (Pinchuk, dim 2, nonconstant Jacobian)**. For CONSTANT Jacobian, everything's open. Unless... 

WAIT. I think I finally remember the right thing: **there IS a known counterexample to the "Jacobian Conjecture" over R^2 with det JF = 1? NO...** Hmm.

What about this: maybe the target is a **Keller map that is injective nowhere... ** no.

Let me try to just SOLVE the algebraic system and see if a solution exists. That's the honest computational approach the user asked for. Set up:

t := xy - 1 (so e = -t = 1 - xy, constant term 1, one other monomial, flip-invariant ✓).

Actually the hint says "a low-degree expression invariant under (x,y) -> (-x,-y). It has constant term 1 and contains only one other monomial." So e(x,y) = 1 + c x^a y^b, a+b even, and "the construction repeatedly uses" e. Most natural: e = 1 - xy or 1 + xy (degree 2). Could also be 1 + x^2 (degree 2! invariant: (-x)^2 = x^2 ✓). Hmm, 1 + x^2 is also plausible! And Pinchuk-type: t = xy - 1...

Let me look at degrees again: (7, 6, 4). If basic blocks are x (deg1), y(deg1), e(deg2):
- deg 7 = 1 + 3·2: x · (deg-3 poly in e)? deg-3 in e means e^3 has deg 6, times x = 7 ✓. So F1 could be x·p(e) + z·q(e) with deg q ≤ 3 in e... or y·..., or x^7, etc.
- deg 6 = 3·2: poly of deg 3 in e, or x^2·(deg 2 in e) = 2+4 = 6 ✓.
- deg 4 = 2·2: poly of deg 2 in e, or x^2·e... 

So plausibly everything is built from x, y, e with z affine. This matches Pinchuk-style "everything from t".

Now let me design the system to solve. I'll use the z-affine ansatz and conditions:
(a) T3 := A1 J(A2,A3) + A2 J(A3,A1) + A3 J(A1,A2) = 0
(b) T1 := A1 J(B2,B3) + A2 J(B3,B1) + A3 J(B1,B2) = 1

Note the BEAUTIFUL symmetry: (a) and (b) are the same functional applied to (A's) and (B's): define Θ(U1,U2,U3; V1,V2,V3) = U1 J(V2,V3) + U2 J(V3,V1) + U3 J(V1,V2). Conditions: Θ(A;A) = 0, Θ(A;B) = 1.

Geometric meaning: for fixed (x,y), Θ(A;B) = 1 is a "volume" condition.

Non-injectivity: I'll look for collisions of flip-related points: F(x,y,z) = F(-x,-y,w). With A_i even under flip and B_i odd: condition A_i(z-w) + 2 B_i = 0 ∀i, i.e., B(x,y) parallel to A(x,y).

So: assume A_i even in (x,y) (i.e., polynomials in e? no — even under flip means monomials x^a y^b with a+b even; polys in t=xy qualify, also x^2, y^2, x^2 y^2...) and B_i odd under flip (a+b odd: x, y, x^3, x^2 y... x^2 y has a+b = 3 odd ✓ — note x^2 y = x·t: "odd" ✓).

Great: take A_i = a_i(t) (polys in t = xy, or in e = 1-xy — same thing), B_i = x P_i(t) + y Q_i(t)? Hmm wait B odd: general odd poly = x·(even) + y·(even) where even = poly in x^2, y^2, xy... but if I restrict "even part" to polys in t only, B_i = x P_i(t) + y Q_i(t). 

Then condition (a) automatic (A_i all functions of t ⇒ all J(A_i,A_j) = 0 ⇒ Θ(A;A) = 0 ✓✓).

Condition (b): computed above for t = xy (I should redo for t = xy - 1, but J's only see derivatives w.r.t. x,y: dt = y dx + x dy either way since constant shifts don't matter! J(f(t), g) with t = xy - 1 vs t = xy: same because ∂t same). 

From my earlier computation with t = xy:
J(B_i, B_j) = x^2 W(P_i,P_j) - y^2 W(Q_i,Q_j) + t^2 (Q_i' P_j' - P_i' Q_j') + (P_i Q_j - P_j Q_i)

Wait, I need to double-check the t^2 term sign and the x^2/y^2 terms. Let me redo this computation cleanly with sympy later. But roughly, condition (b) becomes:

Θ(A;B) = Σ_cyclic a_i(t) [x^2 W(P_j,P_k) - y^2 W(Q_j,Q_k) + t^2(...) + D(P_j,P_k)] = 1

= x^2 Σ a_i W(P_j,P_k) - y^2 Σ a_i W(Q_j,Q_k) + t^2 Σ a_i (Q_j'P_k' - P_j'Q_k') + Σ a_i (P_j Q_k - P_k Q_j)

For this to equal 1 (as polynomial in x, y with x^2, y^2, t^2 = (xy)^2 independent monomial-ish directions — well t^2 = x^2 y^2):

- Σ a_i W(P_j,P_k) = 0 (coeff of x^2) — but careful: is the decomposition unique? A polynomial in x, y can mix: x^2 f(t), y^2 g(t), t^2 h(t), k(t) — monomials x^{2+a} y^a, x^a y^{2+a}, x^{2+a}y^{2+a}, x^a y^a — these are disjoint sets of monomials! ✓ So the four conditions are independent:

(C1) Σ_cyc a_i W(P_j, P_k) = 0
(C2) Σ_cyc a_i W(Q_j, Q_k) = 0
(C3) Σ_cyc a_i (Q_j' P_k' - P_j' Q_k') = 0
(C4) Σ_cyc a_i (P_j Q_k - P_k Q_j) = 1

Beautiful! And these are the four 3×3 determinants: det[a; P; P'] = 0, det[a; Q; Q'] = 0, det[a; Q'; P'] = 0, det[a; P; Q] = 1 — WAIT signs: Σ_cyc a_i W(P_j,P_k) = a1 W(P2,P3) + a2 W(P3,P1) + a3 W(P1,P2) = det[a; P; P'] (rows a, P, P')? det of matrix with rows (a1,a2,a3), (P1,P2,P3), (P1',P2',P3') = a1(P2 P3' - P3 P2') - a2(P1 P3' - P3 P1') + a3(P1 P2' - P2 P1') = a1 W(P2,P3) + a2 W(P3,P1) + a3 W(P1,P2). ✓ YES.

So the whole thing reduces to ONE-VARIABLE polynomial identities:

(C1) det[a; P; P'] = 0
(C2) det[a; Q; Q'] = 0
(C3) det[a; Q'; P'] = 0
(C4) det[a; P; Q] = 1

where a, P, Q are triples of univariate polynomials in t, ' = d/dt.

Interpretation: think of t -> (a(t), P(t), Q(t)) hmm, or better: consider the 3-dim vector spaces. (C1): a, P, P' linearly dependent (over C(t)) — with a, P indep (needed for C4 ≠ 0), this says P' ∈ span{a, P}. Similarly Q' ∈ span{a, Q}. (C3): Q', P', a dependent.

Now: P' ∈ span_{C(t)}(a, P). Differentiate the relation... Let's parametrize: P' = α a + β P, α,β ∈ C(t). Similarly Q' = γ a + δ Q. Then (C3): det[a; Q'; P'] = det[a; γ a + δ Q; α a + β P] = γβ det[a;a;P]... = δα det[a; Q; a]... let me expand: det[a; γa; αa] + det[a; γa; βP] + det[a; δQ; αa] + det[a; δQ; βP] = 0 + γβ det[a;a;P] + δα det[a;Q;a] + δβ det[a;Q;P] = 0 + 0 + δα det[a;Q;a] - δβ det[a;P;Q] = -δα det[a; a; Q]... det[a;Q;a] = 0 (two equal rows). So = -δβ det[a;P;Q] = -δβ·1 (by C4). So (C3) ⇔ δβ = 0 (given C4, αγ whatever). Wait also the term det[a; γ a; β P] = γβ det[a;a;P] = 0 ✓. So (C3) ⇔ δβ = 0.

So either β = 0 (P' ∈ span(a)) or δ = 0 (Q' ∈ span(a)).

Case 1: β = 0: P' = α a, i.e., P_i' = α a_i for all i — so P_i = c_i + ∫ α a_i. If α is a polynomial (and we want P polynomial), fine. Then (C4): det[a; P; Q] = 1 with P_i = p_i^0 + I_i, I_i' = α a_i.

det[a; P; Q] = det[a; p^0; Q] + det[a; I; Q]. Hmm. Take Q' = γ a + δ Q. Then det[a; I; Q]: differentiate? Consider d/dt det[a; P; Q] = det[a'; P; Q] + det[a; P'; Q] + det[a; P; Q'] = det[a'; P; Q] + det[a; αa; Q] + det[a; P; γa + δQ] = det[a';P;Q] + α det[a;a;Q] + γ det[a;P;a] + δ det[a;P;Q] = det[a'; P; Q] + δ·1.

For (C4) = const = 1, need det[a'; P; Q] + δ = 0. Hmm, since det[a';P;Q] is some function, need δ = -det[a';P;Q], which is determined. But ALSO δ appeared... circular but consistent: we get one functional equation. This is solvable territory but let me look for the simplest instance.

SIMPLEST ATTEMPT: constant triples + one moving. Suppose a = (a1, a2, a3) CONSTANT vector? But a_i are polys in t; if constant, A_i constant: F_i = c_i z + B_i: then F is affine in z with constant A — fine but then degrees: F_i = A_i z + B_i(x,y), A_i const ⇒ deg F_i = deg B_i. Degrees (7,6,4) would need B's of those degrees; also if A constant, is F still possibly non-injective? Collision condition: B(x,y) parallel to A = const vector: B(x,y) = λ c. Plausible. But wait: with A constant, the map (x,y,z) -> F: for fixed z... Actually if A3 = 0 const, then F3 = B3(x,y) and surjectivity onto each fiber... hmm, JC-relevant? Let's check condition (C1)-(C4) with a const: (C1) det[a;P;P'] = 0 auto? No: det with first row constant ≠ auto 0. Hmm, C1: a·(P × P')... not auto.

Let me instead try: P = a (componentwise). Then (C1) det[a;a;a'] = 0 ✓ auto. (C4): det[a; a; Q] = 0 ✗. Dead.

Try Q = a: (C2) ✓ auto; (C4): det[a; P; a] = 0 ✗. Dead.

Try a = P + something... Let me try the ODE case β = 0 with α constant, simplest α = 1: P' = a, i.e., P_i = ∫ a_i. And take δ = 0 too (then (C3) ✓ auto): Q' = γ a, γ ∈ C(t). Simplest γ const = c: Q_i = c ∫ a_i + q_i^0 = c P_i + (q_i^0 - c p_i^0) — write Q = c P + r, r constant vector. Then (C4): det[a; P; Q] = det[a; P; cP + r] = det[a; P; r] = 1 (constant!). 

So need: det[a(t); P(t); r] = 1 where P' = a (i.e., a = P'), r a constant vector. det[P'; P; r] = 1. Note d/dt det[P'; P; r] = det[P''; P; r] + det[P'; P'; r] = det[P''; P; r]. So consistency: need det[P''; P; r] = 0.

det[P'; P; r] = r · (P' × P) (scalar triple product) = Σ r_i (P'_j P_k - P'_k P_j) = -Σ r_i W(P_j, P_k). So condition: Σ_cyc r_1 W(P_2, P_3) = -1 (constant), and consistency det[P'';P;r] = Σ r_i (P''_j P_k - ...) hmm = 0.

Simplify: choose r = (0, 0, r3): then condition = r3 (P1' P2 - P2' P1)·... = r3 (P_2' P_1 - P_1' P_2)·(-1)... compute: r·(P' × P) = r3 (P'×P)_3 = r3 (P1' P2 - P2' P1). Set = 1: r3 (P1' P2 - P1 P2') = 1, i.e., W(P2, P1)·r3 = 1: a Wronskian of two univariate polys equal to a NONZERO CONSTANT. 

Wronskian of two polynomials constant: W(P1, P2) = P1 P2' - P1' P2 = const. Classic: e.g., P1 = t, P2 = 1: W = t·0 - 1·1 = -1 ✓. But deg... P2 = 1 constant ⇒ A2 = a2(t) = P2' = 0: F2 = 0·z + B2: A2 = 0. Then F2 = B2 = x P2(t) + y Q2(t) = x·1 + y(c·1 + r2) = x(1) + y(1 + c)... wait r = (0,0,r3) so r2 = 0: Q2 = c P2 + r2 = c. B2 = x·1 + y·c. Hmm F2 = x + c y: degree 1?! But hint says deg F2 = 6. Hmm. But maybe other components carry degree; hint says output order (7,6,4) specifically. So this simplest solution gives wrong degrees but RIGHT STRUCTURE. The real example must use higher-degree P's.

Consistency condition: det[P''; P; r] = r3 (P1'' P2 - P1 P2'')·... = r3 (P_1'' P_2 - P_2'' P_1) hmm sign: (P'' × P)_3 = P1'' P2 - P2'' P1. Need = 0. With P2 = 1: P1'' ·1 - 0 = P1'' = 0 ⇒ P1 linear: P1 = t (up to const). So the (r along 3rd axis, α=δ=0, β=γ=0...) branch forces: P1 = t, P2 = 1, P3 arbitrary!, a = P' = (1, 0, P3'). Then Q = c P + r: Q1 = c t + 0 = c t, Q2 = c, Q3 = c P3 + r3. And W(P2,P1)·r3: W(1, t) = 1·1 - 0·t = 1; set r3 = 1.

So: a = (1, 0, P3'(t)), P = (t, 1, P3(t)), Q = (c t, c, c P3 + 1). F3 = A3 z + B3 = P3'(t) z + x P3(t) + y (c P3(t) + 1). F1 = z + x t + y c t. F2 = x + c y.

Let me VERIFY det JF = 1 for this and check degrees! With c a free constant and P3(t) a free polynomial, t = xy - 1.

F1 = x t + c t y + z = x(xy-1) + c(xy-1)y + z = x^2 y - x + c x y^2 - c y + z. deg 3. Hmm deg 3 not 7. But with P3 of degree d in t, F3 ~ deg 2d+1. To match (7,6,4) we'd need a fancier branch, but let me first CHECK this mechanism actually gives det = 1 and non-injectivity, then upgrade degrees.

Actually hold on, I should double check (C2): det[a; Q; Q'] = 0. Q = cP + r, Q' = c a. det[a; cP + r; c a] = c det[a; cP + r; a] = 0 ✓. (C3): det[a; Q'; P'] = det[a; ca; a] = 0 ✓. (C1): det[a; P; P'] = det[a; P; a] = 0 ✓. (C4): det[a; P; Q] = det[a; P; r] = det[(1,0,P3'); (t,1,P3); (0,0,1)] = expand along third column... = 1·det[[1,0],[t,1]] = 1 ✓✓✓.

WOW. So det JF = 1 (assuming my expansion of Θ and the z-affine computation are right — MUST verify computationally). And this works for ANY polynomial P3(t) and constant c. The degrees: F1 deg 3 (fixed), F2 deg 1, F3 deg = deg(x P3(t)) = 1 + 2 deg P3. To get degrees (7,6,4) we need a richer branch — e.g., nonzero β or δ, or r not axis-aligned, or P, Q both nontrivial with Wronskian constant — but FIRST let me verify the mechanism, including NON-INJECTIVITY.

Non-injectivity: A = (1, 0, P3'), B = (x t + c t y, x + c y, x P3 + y(c P3 + 1)). A even in (x,y)? A depends only on t ✓ even. B odd? x t: x·(xy-1): odd+odd? x·xy = x^2 y (deg 3, odd ✓), x·1 odd ✓. Yes all terms of B odd under flip ✓.

Collision condition: F(x,y,z) = F(-x,-y,w) ⇔ A(z - w) = -2B ⇔ B parallel to A (then w = z + 2λ, B = λ A).

B = λ A: equations: x t + c t y = λ·1; x + c y = λ·0 = 0; x P3 + y(c P3 + 1) = λ P3'(t).

From eq2: x = -c y. Sub into eq1: -c y t + c t y = 0 = λ. So λ = 0! Then eq3: x P3 + y (c P3 + 1) = 0 with x = -c y: -c y P3 + c y P3 + y = y = 0 ⇒ x = 0. But then t = -1 and the point is (0,0): flip fixes (0,0)! F(0,0,z) = (z, 0, y... F(0,0,z) = (z, 0, 0·P3 + 0) = (z,0,0): F(0,0,z) = F(0,0,w) ⇔ z = w: no collision. Hmm! So this particular branch is injective after all?? The parallel locus is just {x = y = 0} where flip has a fixed point. Damn — so close!

So the simple branch threads det=1 but the non-injectivity fails: the "B parallel A" locus collapses to the flip-fixed point. The REAL example must arrange the parallel locus to contain a NON-fixed point. That's where the extra degrees of freedom (and degrees 7,6,4) come in: need B = λ A to have a solution with (x,y) =/= (0,0).

Let me reconsider: with more general P, Q, R (I'll allow B_i = x P_i + y Q_i, A_i = a_i, general satisfying C1–C4), the parallel condition B = λA is 3 equations in (x, y, λ):
x P_i(t) + y Q_i(t) = λ a_i(t), i = 1,2,3.

Eliminate λ: x(P_i a_j - P_j a_i) + y (Q_i a_j - Q_j a_i) = 0 for pairs (i,j). Define R_ij = P_i a_j - P_j a_i, S_ij = Q_i a_j - Q_j a_i (polys in t). Need x R_ij + y S_ij = 0, i.e., (x,y) ∝ (S_ij, -R_ij)... consistent across pairs: need x R_12 + y S_12 = 0 and x R_13 + y S_13 = 0 with (x,y) =/= (0,0), and t = xy - 1 must equal the specific value... Note R, S are functions of t = xy - 1. So: need (x,y) with x/y = -S_12(t)/R_12(t) = -S_13(t)/R_13(t) and xy = t + 1. I.e., x^2 = -(t+1) S_12(t)/R_12(t)·... this is getting complicated, but the point stands: solutions exist iff certain polynomial equations in t have roots — generically they DO over C, and the construction art is to keep det = 1 while arranging it.

Actually hmm, wait. Let me reconsider. Maybe think about it as: B = λA has a 1-parameter family chance... Actually the cleanest: pick a value t0 and require the two vectors (P_i(t0)), (Q_i(t0)), (a_i(t0)) to satisfy: x P + y Q = λ a. For each (x,y) with xy - 1 = t0 (1-parameter family!), need (P,Q,a) at t0 such that the system has a solution (x,y,λ) — overdetermined (3 eqns, 3 unknowns (x,y,λ) but constrained xy = t0+1...). 

Let me now think about what the REAL published example is, because I want to anchor. I'm now ~convinced this is the **Pinchuk-type map in C^3 from the paper by ... "A new look at the Jacobian conjecture"?** or the example in the paper **"Pinchuk maps and the Jacobian conjecture"**... OR — now I really do recall — there's a 2015+ arXiv note claiming a JC counterexample in C^3 with degrees around 7 that was later retracted? Hmm. There was the famous 2004 claim by **S. Orevkov**? No. There was a claimed proof of JC by ... many. A claimed COUNTEREXAMPLE: "A counterexample to the Jacobian conjecture" — I recall one by **T. T. Moh**? no. By **"Nayak"**? hmm. Honestly can't recall a specific (7,6,4) paper.

Doesn't matter. The user's framing: Fable found one; I must find the reasoning. The hints gave me: z-affine, degrees (7,6,4), flip-invariant e = 1 ± xy-ish. My derivation above shows the EXACT algebraic skeleton: z-affine + det=1 ⇔ (C0: A even-ness etc.) the four determinant identities, and non-injectivity ⇔ parallel locus avoiding the fixed point. Now I should:

1. Verify my general derivation computationally (sympy): det JF = T3 z^2... wait earlier: det JF = z^3·0? no: det JF = z^2 T2 + z T1 + T0 with T2 = 0 auto, T0 = 0 auto, T1 = Θ(A;B)... hold on, earlier I found T0 = det(β1,β2,β3) = 0?! But then det JF = z·T1 + z^2 T2 + z^3 T3 with T1 = Θ(A;B)?? Then det JF = z·[...] which vanishes at z = 0!! That contradicts det JF = 1 unless I messed up. Let me recheck: β_i = (B_ix, B_iy, 0)? Row i of JF: (∂_x F_i, ∂_y F_i, ∂_z F_i) = (A_ix z + B_ix, A_iy z + B_iy, A_i). β_i = (B_ix, B_iy, 0). T0 = det[β1;β2;β3]: third column all zero ⇒ 0. T1 = det[α1;β2;β3] + det[β1;α2;β3] + det[β1;β2;α3] where α_i = (A_ix, A_iy, A_i). det[α1;β2;β3]: third column = (A1, 0, 0)^T ⇒ = A1·(B2x B3y - B2y B3x) = A1 J(B2,B3) ✓. Similarly others ⇒ T1 = Θ(A;B) ✓. T2 = det[α1;α2;β3]+det[α1;β2;α3]+det[β1;α2;α3]: each has a row with 0 in third column but other rows have A_i in third column: e.g. det[α1;α2;β3]: third column (A1, A2, 0): = -A1·det[[A2x,A2y],[B3x,B3y]]·... cofactor expansion: = A1 (A2x B3y - A2y B3x)·(-1)^{1+3} + A2 (A1x B3y - A1y B3x)(-1)^{2+3} = A1 J(A2,B3) - A2 J(A1,B3). Hmm not zero! I made an error earlier ("third column of β3 is 0 ⇒ det = 0" — WRONG: only ONE row has zero third entry, determinant isn't zero). Let me redo T2:

T2 = det[α1;α2;β3] + det[α1;β2;α3] + det[β1;α2;α3]
det[α1;α2;β3] = A1 (A2x B3y - A2y B3x) - A2 (A1x B3y - A1y B3x) + 0 = A1 J(A2, B3) - A2 J(A1, B3).
det[α1;β2;α3] = A1 (B2x A3y - B2y A3x) + A3 (A1x B2y - A1y B2x)·(sign: third column (A1, 0, A3): entries at rows 1,3: = A1·M13·(+) + A3·M33·(+)? (-1)^{3+3} = +1: M33 = det[[A1x,A1y],[B2x,B2y]] = A1x B2y - A1y B2x. And M13 = det[[B2x,B2y],[A3x,A3y]]?? no: deleting row1 col3: rows 2,3 cols 1,2: [[B2x, B2y],[A3x, A3y]]: = B2x A3y - B2y A3x. So det = A1 (B2x A3y - B2y A3x) + A3 (A1x B2y - A1y B2x) = A1 J(B2,A3) + A3 J(A1,B2).
det[β1;α2;α3]: third col (0, A2, A3): = A2·(-1)^{2+3} M23 + A3 M33: M23 = det[[B1x,B1y],[A3x,A3y]] = B1x A3y - B1y A3x; M33 = det[[B1x,B1y],[A2x,A2y]] = J(B1,B2)·... = B1x A2y - B1y A2x. det = -A2 J(B1,A3) + A3 J(B1,A2) = A2 J(A3,B1) + A3 J(B1,A2).

T2 = A1[J(A2,B3) + J(B2,A3)] + A2[-J(A1,B3) + J(A3,B1)] + A3[J(A1,B2) + J(B1,A2)]
= A1 J(A2,B3) + A1 J(B2,A3) + A2 J(B3,A1)... (since -J(A1,B3) = J(B3,A1)) + A2 J(A3,B1) + A3 J(A1,B2) + A3 J(B1,A2).

Note J(A1,B2) + J(B1,A2) = J(A1,B2) - J(A2,B1). Hmm, can this simplify if A_i = a_i(t)?? With A's functions of t only: J(A_i, B_j) = a_i'(t) J(t, B_j). So T2 = a1' a2' ... no wait: J(A2,B3) = a2' J(t, B3); J(B2, A3) = a3' J(B2, t) = -a3' J(t, B2). So T2 = a1'[a2' J(t,B3) - a3' J(t,B2)] + a2'[a3' J(t,B1) - a1' J(t,B3)] + a3'[a1' J(t,B2) - a2' J(t,B1)]. Collect J(t,B3): a1' a2' - a2' a1' = 0 ✓. Similarly all cancel!! T2 = 0 when all A_i are functions of the same t. 

And T3 = det[α1;α2;α3] = Θ(A;A) = 0 when A_i functions of t (all J(A_i,A_j) = 0) ✓. 

So det JF = z·T1 + 0 + 0?? NO WAIT — T1 multiplies z^1: det JF = T0 + z T1 + z^2 T2 + z^3 T3 = 0 + z·Θ(A;B) + 0 + 0 = z·Θ(A;B). That VANISHES at z=0! That can't be — did I mislabel? OH NO. I see my error: T1 = Θ(A;B) should be the z-coefficient, but then det JF(0) = T0 = 0 always — contradiction with my sympy-checkable example above where I computed C4 ⇒ det = 1?? Hmm wait, in my example above I "verified" C4 = 1 and claimed det JF = 1, but per T0 = 0, det JF = z·Θ(A;B) = z·1 = z =/= 1!! 

Hold on, something's off. T0 = det[β1;β2;β3] with β_i = (B_ix, B_iy, 0): the matrix is [[B1x,B1y,0],[B2x,B2y,0],[B3x,B3y,0]] — third column is (0,0,0)^T — YES det = 0. So det JF has no constant term?! But det JF should equal... let me test with the IDENTITY map: F = (x, y, z): A1 = 0? F1 = x: A1 = 0, B1 = x. JF = I, det = 1. But T0 formula: β1 = (1, 0, 0), β2 = (0,1,0), β3 = (0,0,0): det = 0. T1 = Θ(A;B) = A1 J(B2,B3) + ... = 0 (all A = 0)... but actual det JF = 1. CONTRADICTION. So my row-splitting is wrong.

Where? Row i = (A_ix z + B_ix, A_iy z + B_iy, A_i). For identity F1 = x: A1 = 0, B1 = x: row = (1, 0, 0) ✓. The multilinear expansion of det over rows z α_i + β_i: T0 = det[β1;β2;β3]. For identity: β3 = (B3x, B3y, 0) = (∂x z, ∂y z, 0)?? WAIT. F3 = z: A3 = 1, B3 = 0. β3 = (B3x, B3y, B3-? ...) — I defined β_i = (B_ix, B_iy, 0) — the third entry of row i is ∂_z F_i = A_i, which belongs to α_i, so β_i's third entry is 0 ✓. For identity: β1 = (1,0,0), β2 = (0,1,0), β3 = (0,0,0); α3 = (0,0,1). det JF = det[β1; β2; α3] = 1 ✓ and that's part of T1 (the z^1 coefficient): det[β1;β2;α3] = A3 J(B1,B2) = 1·1 = 1 ✓. And Θ(A;B) = A1 J(B2,B3) + A2 J(B3,B1) + A3 J(B1,B2) = 0 + 0 + 1 = 1 ✓. So det JF = z T1 + T0: for identity that's z·1 + 0 = z?!?! But det JI = 1, not z!

AH, I see — multilinearity: the z^0 term is T0 = det[β1;β2;β3] only if EVERY row takes the β choice. For identity, rows 1,2 take β (z-degree 0) and row 3 takes z α3?? NO — for identity F3 = z: row 3 = (0,0,1) = α3, and the z in "z α_i + β_i": row i = z·α_i + β_i — for F3 = z: A3 = 1: z·(A3x, A3y, A3) + (B3x, B3y, 0) = z(0,0,1) + (0,0,0) = (0,0,z)?? But the actual third row of JF is (∂x F3, ∂y F3, ∂z F3) = (0, 0, 1). MISMATCH: (0,0,z) vs (0,0,1). 

MY ERROR: row i of JF = (∂x F_i, ∂y F_i, ∂z F_i) = (A_ix z + B_ix, A_iy z + B_iy, A_i) — the third entry A_i has NO z. So row i = z·(A_ix, A_iy, 0) + (B_ix, B_iy, A_i). I mislabeled which entries carry z! Correct: α_i = (A_ix, A_iy, 0) [z-coefficient], β_i = (B_ix, B_iy, A_i) [constant]. 

Redo: T0 = det[β1;β2;β3] = det of [[B1x,B1y,A1],[B2x,B2y,A2],[B3x,B3y,A3]] = A1(B2x B3y - B2y B3x)·(+) - A2(B1x B3y - B1y B3x) + A3(B1x B2y - B1y B2x)·... expanding along third column: = A1 J(B2,B3) - A2 J(B1,B3) + A3 J(B1,B2) = Θ(A;B) ✓. T0 = Θ(A;B) — the constant term! And T1 = det[α1;β2;β3] + ... each α has 0 in third column: det[α1;β2;β3] = third column (0, A2, A3)^T... rows α1 = (A1x,A1y,0), β2 = (B2x,B2y,A2), β3 = (B3x,B3y,A3): expand along col 3: -A2 det[[A1x,A1y],[B3x,B3y]] + A3 det[[A1x,A1y],[B2x,B2y]] = -A2 J(A1,B3) + A3 J(A1,B2) = A2 J(B3,A1) + A3 J(A1,B2). Similarly det[β1;α2;β3] = A1 J(A2,B3)·? : col3 = (A1, 0, A3): = A1 det[[A2x,A2y],[B3x,B3y]] + A3 det[[B1x,B1y],[A2x,A2y]] = A1 J(A2,B3) + A3 J(B1,A2). And det[β1;β2;α3] = A1 J(B2,A3)·?: col3 = (A1,A2,0): = A1 det[[B2x,B2y],[A3x,A3y]] - A2 det[[B1x,B1y],[A3x,A3y]] = A1 J(B2,A3) - A2 J(B1,A3) = A1 J(B2,A3) + A2 J(A3,B1).

T1 = A1[J(A2,B3) + J(B2,A3)] + A2[J(B3,A1) + J(A3,B1)] + A3[J(A1,B2) + J(B1,A2)].

With A_i = a_i(t): J(A_i, B_j) = a_i' J(t, B_j). T1 = a1'[a2' J(t,B3)·... collect terms in J(t, B_k): J(t,B3): from A1 J(A2,B3): +a1' a2'; from A2 J(B3,A1): -A2 J(A1,B3)... J(B3,A1) = -J(A1,B3) = -a1' J(t,B3): gives -a2' a1'. Sum: 0 ✓. Similarly for B1: from A2 J(A3,B1) = a2' a3' J(t,B1)... wait J(A3,B1) appears in A2 J(A3,B1)? My collected T1: A2[J(B3,A1) + J(A3,B1)] and A3[J(A1,B2) + J(B1,A2)]: J(t,B1) terms: A2 J(A3,B1) = a2'a3' J(t,B1); A3 J(A1,B2)? no that's J(t,B2). J(t,B1) also from A3 J(B1,A2) = -a2' J(B1,t)·a3'... J(B1,A2) = -J(A2,B1) = -a2' J(t,B1): so A3·(-a2' J(t,B1)): -a2' a3'. Sum with a2'a3': 0 ✓. Similarly B2 cancels. T1 = 0 ✓ when A's are functions of t.

T2 = det[α1;α2;β3] + det[α1;β2;α3] + det[β1;α2;α3]: each has two rows with zero third entry: e.g. det[α1;α2;β3]: col3 = (0,0,A3): = A3 det[[A1x,A1y],[A2x,A2y]] = A3 J(A1,A2) = 0 (functions of t) ✓. All three = 0. T2 = 0 ✓. T3 = det[α1;α2;α3]: third column all zero ⇒ 0 ✓ auto.

GREAT so with A_i = a_i(t): det JF = T0 = Θ(A;B) = A1 J(B2,B3) + A2 J(B3,B1) + A3 J(B1,B2) — EXACTLY condition (b) = 1, and no z terms at all. My earlier analysis stands (I had just mislabeled T0 vs T1, and "T2 auto-zero" was wrong generally but holds under the a_i(t) ansatz). 

So the corrected summary: **Ansatz**: F_i = A_i(t) z + B_i(x,y), t = xy - 1 (or xy), A_i univariate polys, B_i = x P_i(t) + y Q_i(t) (odd under flip). Then:
det JF = Θ(A;B) and the four Wronskian conditions C1–C4 ⇔ Θ(A;B) = 1. Wait — do I still need separate conditions? Θ(A;B) = x^2·(Σx) + y^2·(Σy)... from my earlier expansion: J(B_i,B_j) = x^2 W(P_i,P_j) - y^2 W(Q_i,Q_j) + t^2 (Q_i' P_j' - P_i' Q_j') + (P_i Q_j - P_j Q_i). Hmm wait, earlier I had B_i = x P_i + y Q_i and computed with t = xy. Let me re-derive J(B_i,B_j) carefully — actually let me just verify with sympy. But structurally Θ(A;B) = x^2·[det a;P;P'] - y^2·[det a;Q;Q'] + t^2·[det a;Q';P']·(±) + [det a;P;Q], and setting = 1 gives the four conditions (monomial sets disjoint ✓).

Hmm wait, actually I realize I should double-check the t^2 coefficient sign and the y^2 sign via direct computation. I'll do all of this in sympy. 

THEN: the earlier "simplest solution" P = ∫a etc. gives deg F1 = 3 — too small. For degrees (7,6,4): need higher-degree structure. General solution of C1–C4: think of t -> curve in the space of... Consider the 3-dim space V of... Actually here's a clean way to generate solutions: 

C4: det[a; P; Q] = 1. C1: det[a; P; P'] = 0. C2: det[a; Q; Q'] = 0. C3: det[a; P'; Q'] = 0.

Reformulate: let u(t) = (a(t), P(t), Q(t)) as three vectors in C^3... think of the 3×3 matrix M(t) with ROWS a, P, Q. C4: det M = 1. C1: det[a; P; P'] = 0 — the rows a, P, P' dependent. If a, P independent: P' = α a + β P. C2: Q' = γ a + δ Q (if a, Q indep). C3: det[a; P'; Q'] = det[a; αa + βP; γa + δQ] = αδ det[a;a;Q] + βγ det[a;P;a] + βδ det[a;P;Q] = βδ·1. So C3 ⇔ βδ = 0 (consistent with before; earlier I had P' vs Q' swapped; fine).

Case β = 0: P' = α a (α ∈ C(t), but for polys need α poly dividing... P_i' = α a_i: since P, a polys, α ∈ C(t) with α a_i all poly: OK generically). Then C4: det[a; P; Q] = 1, differentiate: det[a'; P; Q] + det[a; αa; Q] + det[a; P; Q'] = det[a'; P; Q] + det[a; P; γa + δQ] = det[a'; P; Q] + δ det[a;P;Q] = det[a'; P; Q] + δ. Must = 0 (derivative of 1). So δ = -det[a'; P; Q]. Then Q' = γ a + δ Q with δ determined by a, P. Hmm — Q satisfies a linear ODE with coefficients rational in t... 

Simplest consistent subcase: δ = 0 too, i.e., det[a'; P; Q] = 0, and γ free. Then Q' = γ a: Q_i = q_i^0 + ∫ γ a_i. So P_i = p_i^0 + ∫ α a_i, Q_i = q_i^0 + ∫ γ a_i. Then det[a; P; Q]: write P = p0 + I(α), Q = q0 + I(γ): det = det[a; p0; q0] + det[a; p0; I(γ)] + det[a; I(α); q0] + det[a; I(α); I(γ)]. The last: I(α)_i = ∫ α a_i, I(γ)_i = ∫ γ a_i... 

With α, γ CONSTANT: I(α) = α∫a =: α I, I(γ) = γ I: det[a; I; I] = 0: det[a; P; Q] = det[a; p0; q0] + γ det[a; p0; I] + α det[a; I; q0] = det[a; p0; q0] + det[a; γ p0 - α q0; I]. Let r0 = γ p0 - α q0 (constant vector), s0 = p0, c0 = det[a;p0;q0]... wait det[a; p0; q0]: a is a vector of FUNCTIONS; p0, q0 constant vectors: det[a; p0; q0] = a · (p0 × q0) = Σ_i a_i (p0 × q0)_i — a linear combination of the a_i! And det[a; r0; I] = Σ a_i (r0 × I)_i... = -det[a; I; r0]. So:

C4: L(a) + det[a; r0; I] = 1, where L(a) = Σ m_i a_i (m = p0 × q0 constant), I_i = ∫ a_i.

Hmm. And we need det[a'; P; Q] = 0 too (the δ = 0 condition): det[a'; p0 + αI; q0 + γI] = det[a'; p0; q0] + γ det[a'; p0; I] + α det[a'; I; q0] = L(a') + det[a'; r0; I]·(sign: γ det[a';p0;I] + α det[a';I;q0] = det[a'; γp0 - αq0; I] = det[a'; r0; I]). So condition: L(a') + det[a'; r0; I] = 0.

Note d/dt[L(a) + det[a; r0; I]] = L(a') + det[a'; r0; I] + det[a; r0; a]·(I' = a): det[a; r0; a] = 0 ✓ (repeated row). So the two conditions are consistent (second = derivative of first) ✓. 

So GENERAL SOLUTION (Case β = δ = 0, α, γ const): pick any triple of polys a(t), constants m, r0 ∈ C^3 with: Φ(t) := L(a) + det[a; r0; I] ≡ 1, I' = a.

Hmm, when is a sum Σ m_i a_i + Σ_i (r0 × I)_i a_i... det[a; r0; I] = Σ_cyc a_1 (r0_2 I_3 - r0_3 I_2) = Σ a_i (r0_j I_k - r0_k I_j). So Φ = Σ_i a_i [m_i + (r0_j I_k - r0_k I_j)]. With I_i' = a_i. So Φ = Σ m_i I_i' + Σ (r0_j I_k - r0_k I_j) I_i'. = derivative of [Σ m_i I_i + Σ r0_j (I_k I_i) - r0_k (I_j I_i)]... note (r0_j I_k - r0_k I_j) I_i' cyclic sum = derivative of Σ_cyc r0_1 I_2 I_3·?: d/dt [r0_1 I_2 I_3 + ...]: hmm, Σ_cyc a_1 (r0_2 I_3 - r0_3 I_2) = r0_2 a_1 I_3 - r0_3 a_1 I_2 + r0_1 a_2 I_3·... let me just say Φ = Σ m_i I_i' + (stuff)'. So Φ ≡ 1 integrates to: Σ m_i I_i + G(I) = t + C for some function G quadratic in I. Actually: Σ_cyc a_1 (r0_2 I_3 - r0_3 I_2): is this a derivative? d/dt (r0_2 I_1 I_3 - r0_3 I_1 I_2) = r0_2 a_1 I_3 + r0_2 I_1 a_3 - r0_3 a_1 I_2 - r0_3 I_1 a_2. Compare with target r0-terms in Φ: a_1 (r0_2 I_3 - r0_3 I_2) + a_2 (r0_3 I_1 - r0_1 I_3) + a_3 (r0_1 I_2 - r0_2 I_1). The derivative above covers (r0_2 a_1 I_3 - r0_3 a_1 I_2) + (r0_2 I_1 a_3 - r0_3 I_1 a_2) = matches a_3(-r0_2 I_1)?? target has a_3 (r0_1 I_2 - r0_2 I_1): the -r0_2 I_1 a_3 matches ✓ but target's r0_1 I_2 a_3 vs derivative's ... and target a_2 (r0_3 I_1 - r0_1 I_3) vs derivative -r0_3 I_1 a_2: only HALF matches. So Φ = Σ m_i I_i' + (1/2) d/dt[Σ_cyc r0_1 (I_2 I_3 - I_3 I_2)?? no...

Let me just directly integrate term by term: Φ = Σ_i m_i a_i + Σ_cyc a_1 r0_2 I_3 - a_1 r0_3 I_2. Take the a_1 r0_2 I_3 - a_3 r0_2 I_1 = r0_2 (a_1 I_3 - a_3 I_1) = r0_2 (I_1' I_3 - I_3' I_1) = r0_2 d/dt(I_1 I_3)·? d/dt(I_1 I_3) = a_1 I_3 + I_1 a_3: so I_1' I_3 - I_3' I_1 = a_1 I_3 - a_3 I_1 — NOT a pure derivative (it's like a Wronskian). Hmm right, these are Wronskian-like: W(I_3, I_1)·... 

OK this branch analysis is getting heavy. Let me now simplify by choosing r0 = 0 (i.e., γ p0 = α q0): then Φ = L(a) = Σ m_i a_i = m · a ≡ 1: a LINEAR COMBINATION of the a_i equals 1. THAT's clean! So: C4 (with r0 = 0) ⇔ m·a = 1 for some constant vector m (m = p0 × q0), AND P = p0 + α I, Q = q0 + γ I with γ p0 = α q0.

So: **the skeleton**: choose a(t) = (a1,a2,a3) polys with some constant linear combination = 1 (i.e., Σ m_i a_i(t) ≡ 1 — e.g., a2 = 1 constant and m = (0,1,0)); I = ∫a; P = p0 + α I, Q = q0 + γ I, γ p0 = α q0, and p0 × q0 = m.

Then A_i = a_i(t), B_i = x P_i(t) + y Q_i(t), F_i = A_i z + B_i. det JF = 1 (modulo my sign errors — will verify by sympy). Degrees: F_i = a_i(t) z + x(p0_i + α I_i) + y(q0_i + γ I_i): deg ≈ max(2 deg a_i, 1 + 2 deg I_i) = 1 + 2(deg a_i + 1) = 2 deg a_i + 3 (if α ≠ 0). So deg F_i ∈ {odd}: deg a_i = d ⇒ deg F_i = 2d + 3: d=2 ⇒ 7 ✓!!! F1 deg 7 needs a1 of degree 2. F2 deg 6: even! Doesn't fit 2d+3. Hmm: deg 6 could come from a2(t) z with deg a2 = 3 (deg 6) and α... if the B2-part has lower degree: e.g., p0_2, q0_2 chosen so the I-coefficients cancel... or α q0_2 = ... Actually with m = (0,1,0): a2 ≡ 1: then F2 = z + x(p0_2 + α t·1... I_2 = t (plus const)): deg F2 = 3 (like before). To get deg 6 for F2 with a2 = 1... impossible in this sub-branch (r0 = 0 forces a2 = 1 → deg F2 ≤ 3·...). So the real example must use r0 ≠ 0 or α, γ nonconstant or βδ = 0 with δ = 0, β ≠ 0, etc. Fine — degrees of freedom exist. But rather than solving the general theory by hand, COMPUTATION TIME: I'll parametrize with free coefficients and solve the polynomial equations C1–C4 + degree specs numerically/symbolically in sympy, and then ALSO impose the non-injectivity condition (parallel locus at a non-fixed point), then verify everything.

Hmm wait, but actually — hold on. Let me reconsider the problem. Is it possible the claimed example is genuinely real?? Let me sanity-check against known theory: a (7,6,4) z-affine Keller map in C^3, non-injective. If I actually construct and verify one, that's a disproof of JC — enormous. The algebra above shows the constraints have LOTS of solutions (the C1–C4 system is underdetermined). The non-injectivity needs the parallel-locus condition. I derived a fully verified (once I check signs) FAMILY of Keller maps; the simplest member was injective only because the parallel locus collapsed to the origin. With one more degree of freedom it seems very plausible to get collisions... BUT WAIT — if it were this easy, JC would have been disproved long ago. Where's the hidden obstruction?? Let me think...

Hmm, actually, hmm: let me test the simplest member more carefully for OTHER collisions (not just flip-related). F1 = x t + c t y + z, F2 = x + c y, F3 = P3'(t) z + x P3 + y(c P3 + 1), t = xy - 1. Take c = 0, P3 = 0: F = (x(xy-1) + z, x, y)?? wait B3 = x P3 + y(c P3 + 1) = y: F3 = y·1 = y? B3 = x·0 + y·(0 + 1) = y, A3 = P3' = 0: F3 = y. F = (x t + z, x, y). det JF = 1? F = (x^2 y - x + z, x, y): JF = [[2xy - 1, x^2, 1],[1,0,0],[0,1,0]]: det = expand: (2xy-1)(0·0 - 0·1) - x^2(1·0 - 0·0) + 1(1·1 - 0·0) = 1 ✓!!! Keller! Injective? Given (u,v,w) = F: v = x, w = y, u = v^2 w - v + z ⇒ z = u - v^2 w + v. INJECTIVE, obviously (triangular/tame). Of course — the simplest members are tame automorphisms. The mechanism "B parallel A" never fires. Makes sense.

So the real question: can we make B = λA at some (x,y) =/= origin while keeping det = 1? Note: det = 1 is equivalent to Θ(A;B) = 1, i.e., A1 J(B2,B3) + A2 J(B3,B1) + A3 J(B1,B2) = 1. At a point where B = λA (A =/= 0): Θ = A1 J(B2,B3) + ... hmm at such a point, locally B ≈ λA + ... J(B_i, B_j) at the point: if B = λA exactly at the point (not in a neighborhood), J(B_i,B_j) is about derivatives — no direct contradiction. BUT here's a classical CONSTRAINT I should remember: for a KELLER map, F is a local diffeo — fine — and non-injectivity is precisely what's conjectured impossible. There's no local obstruction; the obstruction (conjecturally) is global/topological. The algebraic system C1–C4 + parallel-locus is a finite-dimensional polynomial system — if it has a solution, that's a counterexample. My derivation says solutions to C1–C4 are plentiful; the parallel condition is extra equations. Whether the combined system has a solution is exactly the JC question in this class. Generically: unknown!

Hmm, hmm, wait — actually now I doubt the whole thing: is it TRUE that any solution of C1–C4 with the parallel property gives non-injective Keller map? The collision needs F(x,y,z) = F(-x,-y,w) with (x,y,z) =/= (-x,-y,w): since flip has no fixed points with (x,y) =/= 0, any parallel point with (x,y) =/= 0 gives a genuine collision. And det JF = 1 everywhere: local diffeo. So YES, that would be a bona fide JC counterexample. 

So the plan is concrete and executable:
1. Implement: t = xy - 1; A_i = a_i(t); B_i = x P_i(t) + y Q_i(t); F_i = A_i z + B_i.
2. Verify (sympy) det JF = Θ(A;B) and expand Θ in the monomial basis to get C1–C4 conditions exactly.
3. Solve C1–C4 with degree constraints (deg F = (7,6,4)) — parametrize a, P, Q with unknown coefficients, use the structural solution (m·a = 1 branch and beyond), get a family.
4. Impose parallel condition: ∃(x,y) =/= (0,0), λ: x P_i + y Q_i = λ a_i — eliminate to polynomial equations on the free parameters; find a solution (numerically first, then exact).
5. Verify final example: det JF ≡ 1 symbolically; exhibit explicit p =/= q, F(p) = F(q), numerically and exactly.

If step 4's system is inconsistent, that "proves" this class can't beat JC — then pivot: different e (1 + x^2?), different B-structure (B_i = x P_i(t) + y Q_i(t) with t = x^2? but t must be flip-invariant with... hint says invariant expression has constant term 1, one other monomial: e ∈ {1 ± xy, 1 ± x^2, 1 ± y^2, 1 ± x^2 y^2, 1 ± x^4, ...}). Also B could include higher odd parts: B_i = x^3 R_i(t) + ... etc.

Also the degrees (7,6,4) hint strongly constrains: in my skeleton, deg F_i = max(2 deg a_i, 2 deg I_i + 1) with deg I = deg a + 1: so deg F_i = 2 deg a_i + 3 (if B-part nontrivial) or 2 deg a_i (if B_i = 0). (7,6,4) = (2·2+3, 2·3+? ...). 7 = 2·2+3 (a1 deg 2, with B1) ✓. 6: even — could be 2·deg a2 with B2 = 0?? But B2 = 0 means F2 = a2(t) z: hmm. Or 6 = 2·3: a2 deg 3, B2 = 0: F2 = a2(t) z, deg 6 ✓. 4 = 2·2: F3 = a3(t) z, deg 4, B3 = 0?? But then F2, F3 both z-only... F = (a1 z + B1, a2 z, a3 z): F(x,y,z) all multiples of z except F1: then F(x,y,0) = (B1, 0, 0): the z = 0 plane maps into a line — massively non-injective (F(x,y,0) = (B1(x,y), 0, 0): varying y at fixed... B1(x,y) = x P1 + y Q1: F(0,y,0) = (y Q1(-1), 0, 0) vs F(x, y', 0)... are there collisions? F(x1,y1,0) = F(x2,y2,0) iff B1(x1,y1) = B1(x2,y2): B1 linear-ish in x,y at fixed t... For (x,y) and (-x,-y): B1 odd ⇒ B1(-x,-y) = -B1(x,y) = B1(x,y) iff B1(x,y) = 0: a 1-dim curve of (x,y) (B1 = x P1(xy-1) + y Q1(xy-1) = 0): on this curve F(x,y,0) = (0,0,0) = F(-x,-y,0)!! COLLISION whenever B1(x,y) = 0 has a nonzero solution — which it does over C (polynomial in... B1 = 0 is one equation in 2 vars: has solutions; nonzero ones exist since B1(x,0) = x P1(-1): if P1(-1) ≠ 0 then x = 0 only... hmm: B1(x,y) = 0: solutions: e.g. x = 0 ⇒ y Q1(-1) = 0: if Q1(-1) ≠ 0: only y = 0. Parameterize: need x P1(t) = -y Q1(t), t = xy - 1. Set y = 1: x P1(x - 1) = -Q1(x - 1): polynomial equation in x: x P1(x-1) + Q1(x-1) = 0: leading term x·(P1 leading)·x^d: degree d+1 ≥ 1: HAS A ROOT x0 over C; if x0 = 0 then Q1(-1) = 0 contradiction w/ assumption... so root x0: if x0 = 0, Q1(-1) = 0, but then take y = 2 instead: generic y gives root ≠ 0 (unless x P1 + y Q1 ≡ 0 as polynomial after substitution... can dodge). SO: if F2 = a2 z, F3 = a3 z exactly (B2 = B3 = 0), then F(x,y,0) = (B1(x,y), 0, 0) and B1(x,y) = 0 off-origin ⇒ F(p) = (0,0,0) = F(q) for p =/= q (p and -p both on zero set... wait need F(x,y,0) = F(-x,-y,0): both = (0,0,0) ✓ since B1 = 0 at (x,y) and B1 odd ⇒ = 0 at (-x,-y) ✓). NON-INJECTIVE! And det JF = 1 by the C-system with B2 = B3 = 0: Θ(A;B) = A2 J(B3,B1) + A3 J(B1,B2) + A1 J(B2,B3) = A2 J(0,B1) + A3 J(B1, 0) + A1 J(0,0) = 0 ≠ 1!!! ✗. Dead — det would be 0. Of course: if B2 = B3 = 0 then J-terms vanish. Right: Θ needs B's. OK.

So degrees (7,6,4) must ALL have B-parts... but 6 and 4 are even: B_i = x P_i + y Q_i has ODD degree only (x·even = odd deg). Hmm! B_i(x,y) odd under flip ⇒ only odd total degrees. A_i(t) z: even degrees. F_i = A_i z + B_i: contains BOTH parities ⇒ deg F_i = max(even part, odd part): 7 = max(2·deg a1, odd part): odd 7 = 1 + 2·3: B1 has degree 7: P1 or Q1 deg 3 in t (deg 6) times x: 7 ✓; and a1 deg ≤ 3 (z-part deg ≤ 6). For F2 deg 6 = max(even ≤ 6, odd ≤ 5): a2(t) deg 3 with z-part = 6, B2 odd deg ≤ 5. For F3 deg 4: a3 deg 2, z-part 4, B3 odd ≤ 3.

Hmm OK. So plausible degree budget:
- F1: B1 deg 7 (P1/Q1 deg 3), a1 deg ≤ 3.
- F2: a2 deg 3 (z-term deg 6), B2 deg ≤ 5.
- F3: a3 deg 2 (z-term deg 4), B3 deg ≤ 3.

That's a nice fit! So the ansatz dimension: a1 (deg ≤ 3, 4 coeffs), a2 (deg 3, 4), a3 (deg 2, 3), P_i, Q_i (P1,Q1 deg ≤ 3 (4 each), P2,Q2 deg ≤ 2 (3 each), P3,Q3 deg ≤ 1 (2 each)): total unknowns ~ 4+4+3 + 8 + 6 + 4 = 29 coefficients. Equations: C1–C4 are polynomial identities in t of various degrees — a bunch of equations. Plus parallel-locus condition. Overdetermined-ish but with structure. sympy/nsolve or exact Gröbner might crack it, especially guided by the skeleton solution (m·a = 1 branch) which I can use to REDUCE unknowns: set a2 = 1?? but deg F2 = 6 needs a2 deg 3... unless B2... wait with m·a = 1: some fixed linear combo of a's is 1; if m = (m1, m2, m3) with all nonzero, a's can individually have high degree. E.g., m1 a1 + m2 a2 + m3 a3 = 1: a1 deg 3, a2 deg 3, a3 deg 2 with high terms canceling in the combination. OK.

And on the skeleton: P = p0 + α I, Q = q0 + γ I, γp0 = αq0: then B_i = x p0_i + y q0_i + I_i(t)(α x + γ y). Write u = α x + γ y (odd linear), v0_i = x p0_i + y q0_i (linear): B_i = v0_i + I_i(t) u. NICE FORM: 

F_i(x,y,z) = a_i(t) z + I_i(t) u + v0_i,   t = xy - 1,   u = α x + γ y,   v0_i = p0_i x + q0_i y,

with constraints: m·a ≡ 1 (m = p0 × q0), γ p0 = α q0.

Wait, but this has deg F_i = 2 deg a_i + 3 always (B-part dominates with odd degree 2deg a+3). So deg pattern would be odd×3 — can't make (7,6,4). UNLESS for i = 2, 3 the B-part cancels: B2 deg < 2 deg a2 + 3 requires I_2(t) u + v0_2 to have smaller degree — impossible unless I_2 = 0 (a2 = 0 ⇒ then a2 = 0, F2 = v0_2 linear). Hmm. So skeleton (with α, γ const, r0 = 0) only yields odd degrees. The even degrees 6, 4 must come from the z-terms being the TOP terms with B-parts smaller: B_i smaller requires... in skeleton B_i = v0_i + I_i u: deg 2(deg a_i)+1 > 2 deg a_i always (when I_i ≠ 0). So for F2: need I_2 = 0 ⇒ a2 = 0 ⇒ but then z-term for F2 is 0, deg F2 = deg B2 = deg v0_2 = 1. Contradiction with deg 6. ⇒ Skeleton too rigid. Need the general case (α(t), or β ≠ 0, or r0 ≠ 0...).

General case again: C1: P' = α a + β P (rational functions allowed). If β ≠ 0 then δ = 0: Q' = γ a. Hmm — then by the earlier consistency (differentiate C4): det[a';P;Q] + det[a;P';Q] + det[a;P;Q'] = det[a';P;Q] + det[a; αa + βP; Q] + det[a;P;γa] = det[a';P;Q] + α det[a;a;Q] + β det[a;P;Q] + γ det[a;P;a] = det[a';P;Q] + β. Need = 0 ⇒ det[a'; P; Q] = -β.

So the GENERAL solution (case β≠0, δ=0): a, P, Q polys, γ, α, β rational with:
(i) P' = α a + β P
(ii) Q' = γ a
(iii) det[a;P;Q] = 1
(iv) det[a';P;Q] = -β (= -β·det[a;P;Q] consistent with (iii) derivative ✓ automatically if (i),(ii),(iii) hold! because d/dt det[a;P;Q] = det[a';P;Q] + det[a;P';Q] + det[a;P;Q'] = det[a';P;Q] + α·0 + β·1 + γ·0 — wait det[a; αa; Q] = α det[a;a;Q] = 0 ✓; det[a; βP; Q] = β·1 ✓; det[a;P;γa] = γ det[a;P;a] = 0 ✓. So (iii) ⇒ need det[a';P;Q] + β = 0. (iv) is NOT automatic but is the integrability companion.)

From (i): a = (P' - β P)/α. Substitute in (iii): det[(P' - βP)/α; P; Q] = (1/α)(det[P';P;Q] - β det[P;P;Q]) = (1/α) det[P';P;Q] = 1 ⇒ det[P'; P; Q] = α. And (iv): det[a';P;Q] = -β.

Hmm, getting complicated. Let me just go computational: parametrize everything with unknowns and throw at a solver. Actually, better idea: since (iii) says rows a, P, Q form a matrix in SL_3(C(t))... 

Hmm, you know what, let me reconsider — maybe I'm overcomplicating. Let me revisit: maybe B_i are NOT of the form x P_i(t) + y Q_i(t). The odd-under-flip polynomials in x,y that are "built from e = 1 - xy": odd poly = x·E1(t) + y·E2(t) is the general odd polynomial ONLY if we restrict to polys in x, y, t with t = xy: indeed any polynomial in x, y can be reduced mod (t - xy): x^a y^b with a ≥ b: x^{a-b} t^b or y^{b-a} t^a: so C[x,y] = C[t] ⊕ x C[t]... no wait: x^a y^b = x^{a-b} (xy)^b if a ≥ b. So basis: {t^b} ∪ {x^k t^b} ∪ {y^k t^b}, k ≥ 1. Odd under flip: x^{2j+1} t^b, y^{2j+1} t^b — so general odd: x·(poly in x^2, t) + y·(poly in y^2, t) — NOT just x P(t) + y Q(t)! There are x^3, x^5, ... terms: x^3 R(t) etc. So my ansatz was too narrow! The full odd space: x·S(x^2, t) + y·T(y^2, t). Hmm, but degrees: (7,6,4): B1 deg 7: could be x^7, x^5 t, x^3 t^2, x t^3, y^7... "x t^3" fits deg 7 with only t. But maybe the real example uses x^3 terms etc.

OK. At this point, hands-on computation is the way. But also — let me reconsider whether the intended example might be something I can find in literature memory: "Jacobian conjecture counterexample C^3 degree 7" — hmm, there's the famous **"unimodular rows"**... no. There's **Suslin**... no. **Vénéreau**: f1 = y + x^2 z + x y^2... wait Vénéreau's polynomial: f = y + x(xz + y^2) + x^2(...)... The Vénéreau coordinate: y + x^2 z + xy^2? deg 3. Berson–Vénéreau... related to A^2-fibrations, not JC counterexamples.

Hmm what about the **"Cornell–... "** no. What about **Miyanishi's**... no.

I also recall the **"Gorni–Zampieri"** construction: they embed Pinchuk-like maps into tame-related maps, "Yagzhev–Pinchuk"... Actually! Gorni–Zampieri: "A Pinchuk map in dimension 3 with constant Jacobian"? Hmm, no... They have a paper "An example of a cubic polynomial map with nilpotent Jacobian..." hmm.

Actually you know what? I bet the intended answer is a REAL object: **Pinchuk-style non-injective Keller map over R^3? or C^3 constructed as F(x,y,z) = (f(x,y) + z e(x,y)·u1...)** hmm. OR the famous example by **Zbigniew Jelonek** of an etale non-injective map C^3 -> C^3?? THAT DOESN'T EXIST (would disprove JC).

Honestly, maybe the Fable counterexample is a real thing in THIS fictional scenario, and my job is just to construct the object per the hints. The algebra says it's feasible-ish. Let me now get concrete with computation. Let me set up sympy, encode the general z-affine ansatz with t = xy - 1, A_i = a_i(t), B_i = x P_i(t) + y Q_i(t) first (narrow), solve C1–C4 with degree budget (7,6,4), then the parallel condition. If narrow fails, widen B to include x^3, y^3 terms.

Hmm wait, before heavy computation, let me reconsider the parallel condition on the skeleton to see how close we get and what freedom

I was mid-derivation. Let me continue with the plan, but break it into smaller pieces: first verify the core algebraic identity with sympy, then solve the system. Let me print my thinking concisely and compute.

Key derived facts to verify computationally:
- F_i = A_i(t) z + B_i(x,y), t = xy - 1, A_i = a_i(t), B_i = x P_i(t) + y Q_i(t)
- det JF = Θ(A;B) = A1 J(B2,B3) + A2 J(B3,B1) + A3 J(B1,B2) (z-terms cancel when A_i are functions of t)
- Θ expands into four monomial sectors giving conditions C1–C4 (determinant identities in t)
- Non-injectivity: collision F(x,y,z) = F(-x,-y,w) iff B(x,y) parallel to A(x,y) at a non-fixed point (x,y) ≠ (0,0).

Let me verify the det identity first with a small sympy script.

My reduction is flawed (it left 1/y terms — the remainder approach divides by x only, producing y-powers in denominators). Let me do a cleaner approach: directly expand Theta in monomials x^i y^j and match against x^2 f1(t) + y^2 f2(t) + t^2 f3(t) + f4(t) via monomial bookkeeping: monomial x^i y^j:
- belongs to sector "x^2 f(t)": i = j + 2
- sector "y^2 f(t)": j = i + 2
- sector "t^2 f(t)": i = j ≥ 2
- sector "f4(t)": i = j
Note overlap: i=j≥2 belongs to both t^2-sector and f4-sector. That's fine — the split isn't unique there, but with the specific formulas D3, D4 it should match as: coefficient of x^i y^i = [t^i](t^2 D3 + D4) = [t^{i-2}]D3 + [t^i]D4.

Let me just compare monomial by monomial.

The sector structure is confirmed (i-j ∈ {-2, 0, 2} only — exactly the predicted four sectors), but my D formulas are off. Let me extract the actual sector polynomials and reverse-engineer the correct determinant formulas. Set up tasks too since this is a multi-step effort.