# Do closed, connected subsets of manifolds always admit open neighborhoods to which they are homotopy equivalent?

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.

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.
• Thank you. Are there any mild conditions we could impose on $C$ to make sure that such a $U$ exists? Dec 24 '20 at 1:44
• 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. Dec 24 '20 at 2:52