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2006.15322

AROUND EFIMOV’S DIFFERENTIAL TEST FOR HOMEOMORPHISM

Victor Alexandrov

correcthigh confidence
Category
math.DS
Journal tier
Specialist/Solid
Processed
Sep 28, 2025, 12:55 AM

Audit review

The uploaded paper is a survey that states Efimov’s 1968 Theorem 2 exactly with the hypotheses |1/a(x) − 1/a(y)| ≤ C1|x−y| + C2 and |det f′(x)| ≥ a(x)|curl f(x)| + a(x)^2 (with det f′ < 0) and concludes that f(R^2) is convex and that f is a homeomorphism onto its image; see the verbatim statement labeled Theorem 2 in the survey . As a survey, it does not re-prove the theorem but correctly documents it and related results (e.g., Theorem 3 and variants) . The candidate solution follows the standard ‘band’ method but commits material errors: (i) a misapplication of Green’s/ Stokes’ theorem leading to an incorrect boundary term for ∫Ω s curl f (they write ∮ s J f·τ rather than the correct ∮ s f·τ), and then assert boundary cancellations that do not follow; (ii) an unjustified claim that the inequality on 1/a controls total variation along curves (the additive C2 prevents a uniform variation bound), and (iii) an incorrect identification of geometric lengths with t2−t1 on the wrong boundary segments. These issues break the core contradiction step, so the model’s proof is not correct, even though the high-level plan mirrors Efimov’s method reported in the paper .

Referee report (LaTeX)

\textbf{Recommendation:} reject

\textbf{Journal Tier:} specialist/solid

\textbf{Justification:}

While the solution follows the classical Efimov blueprint (level-set analysis, quadratic determinant inequality, band argument), it makes crucial technical mistakes: the Green/Stokes step is misapplied and the ensuing boundary cancellations do not hold as claimed; the growth condition on 1/a is misused to deduce a total-variation bound; and the key linear lower bound for the positive s\^2-term is not established. These issues prevent the proof from reaching the advertised contradiction, so the submission does not meet correctness standards.