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A nonempty topological space $X$ is a cut point space if every point $p \in X$ is a cut point. A point $p$ in a space $X$ is a cut point if $X$ is connected and $X \backslash \{p\}$ is disconnected.

Question: Is a locally 1-Euclidean cut point space simply connected?

See Lemma 1 and the text following Lemma 1 from a translation by Sorcar on arxiv of a 1957 paper by Haefliger and Reeb. The Haefliger-Reeb paper takes a baseline assumption that spaces are second countable. So the text following Lemma 1 seems to claim that a second countable + locally 1-Euclidean + cut point space is simply connected; there is no argument though. This question is about asking if second countable is necessary, and asking for a simple proof of the claim without second countability, or asking if a proof of this appears in the literature.

Geoffrey Sangston
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    I don’t think second countability matters. A nonempty, open, connected subspace of a $T_1$ cut point space is always a cut point space itself. So given a loop $f:S^1\to X$, its image is compact. You can thus always cover it by finitely many open sets homeomorphic to $\mathbb{R}$. The union of those neighborhoods is then a second countable, locally 1-Euclidean, cut point space, so simply connected by the result you cite. But this means the original loop is null-homotopic in this smaller space, so of course also null-homotopic in $X$. – David Gao Dec 06 '24 at 20:02
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    (Though I cannot really think of a proof of the cited result with second countability. It certainly looks plausible, though.) – David Gao Dec 06 '24 at 20:05
  • It might be helpful to determine if Theorem 6 from McCord's Singular homology groups and homotopy groups of finite topological spaces applies to the Hausdorff quotient map. Since the Hausdorff quotient map seems to be complicated in general, I'm not sure how to check this though. – Geoffrey Sangston Dec 16 '24 at 16:58

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