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Say $X \subseteq \mathbb R^n$ has a sufficiently nice structure such as being a compact manifold with boundary (I'm more interested in standard spaces than anything else.)

Is it true that if $A \subsetneq\partial X$ is a closed and contractible proper subset of $\partial X$, then $X / A$ is homeomorphic to $X$?

It's clear that the homology agrees by excision, but I'm not sure if its easy in general to see such a homeomorphism. I think that in low dimensions that such a result would be reasonable, but I'm not sure how one would prove such a statement.

I suppose that $2$-manifolds (with boundary) will not change genus, so we can expect this result for manifolds in three dimensions. $1$- manifolds will not change either, which I think is clear.

Andres Mejia
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  • $A$ needs to be closed if $X$ is $T_1$, and compactness is certainly useful if $X$ is first countable. Idenitfying an unbounded closed set could leave us with a non first-countable space. – Henno Brandsma Feb 18 '17 at 09:14
  • I specified that $X$ is a subset of Euclidean space, and $A$ was a closed subset, although that was an interesting point, thank you – Andres Mejia Feb 18 '17 at 10:52
  • http://math.stackexchange.com/questions/621593/contracting-a-contractible-set-in-mathbb-r2/623556#623556 – Moishe Kohan Feb 18 '17 at 18:12

1 Answers1

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The answer is negative in dimensions $\ge 4$. The relevant construction is due to Bing: Start with a wild arc $A$ in $S^3$ such that $\pi_1(S^3-A)\ne 1$. Take the quotient $Y:=S^3/A$. It is not a manifold. Indeed, denoting $a$ the projection of $A$ to $Y$, we notice that $a$ does not have a basis of neighborhoods $U_i$ with $\pi_1(U_i -a)=1$. To obtain an example you are interested in, consider $X=B^4$.

On the other hand, if you take a surface $S$ (maybe with boundary) and contract a compact contractible subset $A\subset S$ to a point, then the quotient is still a surface (this is due to Moore, who proved much more) and hence homeomorphic to $S$. (It suffices to know that $A$ is "cell-like"; you can also take the quotient of $S$ by an upper semicontinuous partition of it into cell-like compact subsets.)

From this it follows that your question has positive answer for $n=1, 2$ and, most likely, $3$ (but the latter requires a bit more thought).

Edit 1: see also this question and references given there.

Edit 2. The statement indeed holds in dimension 3 as well.

Theorem. Suppose $M$ is a 3-dimensional manifold with boundary, $A\subset \partial M$ is a compact contractible subset. Then the quotient space $M/A$ is a manifold with boundary, map $M\to M/A$ is a homotopy equivalence which is homotopic to a homeomorphism.

The proof uses mostly the Dehn Lemma.

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Moishe Kohan
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  • Ah excellent, is there any way to place stronger conditions on the type of subset (finite polygonal chain etc.) to get this going? This is an excellent answer +1. – Andres Mejia Feb 18 '17 at 20:01
  • @AndresMejia: Maybe $A$ a subcomplex in a triangulation does suffice, I will think about it. – Moishe Kohan Feb 18 '17 at 20:05
  • Thank you, I appreciate your interest in the question. If you have a good reference to better understand Dehn's Lemma, I would be glad to take a look as well. – Andres Mejia Feb 18 '17 at 20:09
  • btw this may be a more basic question: what can we say about being homotopic to a homeomorphism? We already know that there exist maps s.t when composed are homotopic to the identity, is the former condition stronger somehow? – Andres Mejia Feb 18 '17 at 20:11
  • @AndresMejia: The hardest part, I think, is showing that the quotient is a topological manifold. It follows from old work of Bob Edwards that if $n$ is reasonably large (say, $n\ge 6$), $M$ is PL and $A\subset M$ is a contractible subcomplex, then $M/A$ is a topological manifold. Not sure about $n=4, 5$. – Moishe Kohan Feb 18 '17 at 20:26