Does there exist a bijection of $\mathbb{R}^n$ to itself such that the forward map is connected but the inverse is not?
candidateconfidence 50%
VibeMathed reports this item as candidate. VibeMath preserves that report as a source assertion and has not independently authored a plain-language mathematical explanation.
Precise statement
Willie Wong asked on MathOverflow in April 2016: if $f:\mathbb R^n\to\mathbb R^n$ is a bijection that maps every connected set to a connected set, must $f^{-1}$ do the same? By Tanaka's theorem and invariance of domain this is equivalent to asking whether every connectedness-preserving bijection of $\mathbb R^n$ is continuous. For $n=1$ the answer is yes. For $n\ge 2$ the question stayed open for a decade: the top-voted answer constructs such a bijection only from $\mathbb R$ to $\mathbb R^2$, and Banakh and Banakh (2020) proved continuity in several compact settings while calling Wong's problem still open.
The source statement is reproduced for indexing with attribution. Mathematical correctness requires domain-expert or mechanical review. VibeMath has not independently audited statement
fidelity, correctness, priority, or novelty.
What AI did
GPT-6 (Codex, Ultra effort), Claude Fable 5.1
The preprint's own disclosure: the paper "is the outcome of a research program conducted with two AI systems under the author's direction: OpenAI's Codex (GPT-6, Ultra effort), which produced the structural theory, the constructions, the adversarial audits, the verification of the argument, and the draft; and Anthropic's Claude Fable 5.1 (Extra effort), which planned the program, reviewed the successive run reports, and proposed the single-line coloring that makes the construction uniform in the dimension." The author chose the problem, wrote the briefs and ran the audits between systems. Appendix A separates ideas taken from the literature from ideas first recorded within the program.
Unreviewed. The 17-page preprint (Zenodo 10.5281/zenodo.22346412, version 1, dated 5 September 2026) was read here on the day it appeared; the theorem, the construction outline and the disclosure match the submission. Nobody outside the author's program has checked the argument, and the preprint is visibly unfinished: its acknowledgements read "to be supplied by the author" and its disclosure ends with a bracketed statement "to be completed after review" that the author has verified the mathematics and accepts responsibility. Candidate until that statement is filled in and someone independent has read the proof. The question's history warrants care: it drew five answers over ten years, all partial, and a 2020 paper by Banakh and Banakh devoted to it.
Recurrent tubes and connectedness-preserving bijections of Euclidean spaces (Zenodo preprint, 5 Sep 2026)
Answered in the negative for every $n\ge 2$. The preprint constructs a bijection $F:\mathbb R^n\to\mathbb R^n$ that maps every connected set to a connected set, is continuous exactly off the closed ray $[0,\infty)\times\{0\}^{n-1}$, and pulls the straight segment $\{(1,0,\dots,0)\}\times[0,1]$ back to the middle-thirds Cantor set on that ray, so $F^{-1}$ is not connectedness-preserving. The construction extends a thin solid tube by finger moves so its cross-sections recur near every point of the complementary compactum, collapses the ray onto the tube's ideal end, and certifies arbitrary connected sets by a separation argument; $F$ and $F^{-1}$ can be taken Borel. The same author's companion note on Darboux injections from closed manifolds (Banakh-Banakh Problems 1.7 and 1.8) is a separate result and belongs in its own entry.
Known method families
construction (source-reported)
Source-reported tools: construction.
Independent: unknown · difference confidence: 0
What remains uncertain
VibeMath has not independently audited the mathematical statement, proof, or novelty claim.
The source status is candidate and must not be represented as solved.
VibeMath has not independently verified the mathematical claim.
AI-attempt independence and training-data exposure are unknown unless explicitly documented.
VibeMath has not independently audited the mathematical statement, proof, or novelty claim.