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For this question a manifold is a locally upper-Euclidean Hausdorff space; paracompactness or second-countability is not assumed, and boundary may be present.

Let $M$ be a manifold. What are sufficient conditions for there to exist open $U \subset M$ such that $\partial M \subset U$ and $U$ (strongly?) deformation retracts onto $\partial M$? (Could this be true for every manifold?)

Clearly $\partial M$ being collared in $M$ is a fairly general sufficient condition (this includes paracompact manifolds). However, I'd be interested in whether there is a more general sufficient condition than this.

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  • $\begingroup$ I can't find a reference to " locally upper-Euclidean"; what does this mean? Do you mean locally homeomorphic to the upper-half space? $\endgroup$
    – ABIM
    Commented Mar 25, 2020 at 9:50
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    $\begingroup$ Yes. For each point there exists an open neighborhood homeomorphic to an open subset of $\mathbb{R}^{n - 1} \times \mathbb{R}_+$, where $\mathbb{R}_+ = \{x \in \mathbb{R} : x \geq 0\}$, i.e. $M$ is a manifold-with-boundary but not necessarily metrizable. $\endgroup$
    – kaba
    Commented Mar 25, 2020 at 13:12
  • $\begingroup$ Thanks kaba, just wanted to be sure I wasn't missing something important :) $\endgroup$
    – ABIM
    Commented Mar 25, 2020 at 13:15
  • $\begingroup$ I do not understand the question: Boundary of every topological manifold is collared. What else would you need? $\endgroup$
    – Moishe Kohan
    Commented Feb 20, 2022 at 7:53
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    $\begingroup$ @MoisheKohan This is not true if one does not assume metrizability. See for instance this question (and the discussion in the comments): mathoverflow.net/questions/344104/… $\endgroup$
    – Mathieu Baillif
    Commented Feb 20, 2022 at 9:10

1 Answer 1

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The following theorem and example give (partial) answers. By an $n$-manifold we understand a Hausdorff space whose any point has a neighborhood, homeomorphic to an open subspace of $\mathbb R^n_+=\{(x_1,\dots,x_n)\in\mathbb R^n:x_n\ge 0\}$.

Theorem. For each 1-manifold $M$ the boundary $\partial M$ is a deformation retract of some open set $U\subseteq M$ that contains $\partial M$.

Proof. By the definition of a 1-manifold, for every $x\in\partial M$ there exists a neighborhood $U_x\subseteq M$, homeomorphic to $[0,1)$. Fix a homeomorphism $h_x:U_x\to[0,1)$ such that $h_x(x)=0$. It can be shown that for distinct $x,y\in\partial M$ the sets $U_x,U_y$ are disjoint. Then $U=\bigcup_{x\in \partial M}U_x$ is an open neighborhood of $\partial M$ and the function $H:U\times [0,1]\to U$, $H:(u,t)=h_x^{-1}(t\cdot h_x(u))$ where $u\in U_x$, is a deformation retraction of $U$ onto $\partial M$.

Example. There exists a separable 2-manifold $M$ whose boundary $\partial M$ is homeomorphic to $\mathbb R\times\mathfrak c$. Being non-separable, the boundary $\partial M$ cannot be a retract of a (necessarily separable) neighborhood $U$ of $\partial M$ in $M$.

This example is described by Peter Nyikos as Example 3.6 of his survey paper "The theory of nonmetrizable manifolds" in the Handbook of Set-Theoretic Topology.

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  • $\begingroup$ Thanks! This goes a long way. $\endgroup$
    – kaba
    Commented Feb 25, 2022 at 21:00
  • $\begingroup$ Unless there will be other answers, I'll probably accept this as an answer; the question has been open for almost two years now. The collar provides a reasonable sufficient condition, and the example here shows that a retract is not always possible. $\endgroup$
    – kaba
    Commented Feb 25, 2022 at 21:04

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