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Let $M$ be a topological manifold and $C$ a closed, connected subset of $M$. Can we always find an open neighborhood $U$ of $C$ such that the inclusion $C \to U$ is a homotopy equivalence?

My thought was to take $U$ as a union of small neighborhoods around each point in $C$ and show that $U$ deformation retracts onto $C$. However, I don't know how to make this rigorous. If we're working with smooth manifolds and $C$ a submanifold, we can just take a disk bundle of the normal bundle. But in general, $C$ may not have such a tubular neighborhood.

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This fails miserably already for $M=\mathbb{R}^2$. The inclusion $C\rightarrow U$ being a homotopy equivalence would imply that $\pi_1(C)\cong\pi_1(U)$, but $U$, being an open subset of $\mathbb{R}^2$, is a noncompact surface and these are known to have free fundamental group (see here). On the other hand, there are plenty of closed, connected subsets of the plane with non-free fundamental group, the most famous one probably being the Hawaiian earring.

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  • $\begingroup$ Thank you. Are there any mild conditions we could impose on $C$ to make sure that such a $U$ exists? $\endgroup$
    – inkievoyd
    Dec 24 '20 at 1:44
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    $\begingroup$ There are some sufficient conditions such as $M$ having a CW structure for which $C$ is a subcomplex, $M$ being smooth and $C$ being an embedded smooth submanifold or $C$ being a locally flat codimension $1$ submanifold, but none of these are mild. I'm not knowledgeable on this topic, but I believe someone with expertise could probably give you a more satisfactory answer, so perhaps this is worth asking as a separate follow-up question. $\endgroup$
    – Thorgott
    Dec 24 '20 at 2:52

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