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To continue this line of thought, say we have a compact, simply connected subset $K$ of a manifold $M$. Can a simply connected neighborhood of $K$ be found? I suppose that if $K$ is a submanifold, this is possible by using the tubular neighborhood theorem (We reduce the problem to finding a simply connected neighborhood of K in it's normal bundle, which is doable). I'm also looking for an elementary proof in the case that $K$ is homeomorphic to the closed interval.

cmk
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Hal
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  • There are examples with $K\cong [0,1]$ if you additionally assume that your neighborhood is a compact manifold with boundary. – Moishe Kohan May 29 '20 at 22:52

1 Answers1

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Not in general.

Consider $M=S^1\times\mathbb{R}$ and let $K\subseteq M$ be the (closed) topologist's sine curve wrapped around the tube and connected to itself. This is a compact, simply connected space, because no loop in $K$ goes around the tube. But every open neighbourhood of $K$ contains a loop around the whole tube, which cannot be contracted.

The question you refer to additionally assumes that $K$ is locally path connected. And this makes the above counterexample invalid. I remember thinking about that question. Unfortunately I still don't know the answer. The nature of your question is almost the same as in the question referred to by the question you referred to :D Simply connected neighbourhood of a simply connected closed set

The only difference is that you assume that $K$ is compact, although pretty much the same counterexample is given there. Funny thing is that it was given by me as well, wow, the world is small.

freakish
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  • Thanks for the answer! I've upvoted but I'm a bit hesitant to accept because at least for something homeomorphic to $[0,1]$, the answer I'm quite invested in – Hal May 27 '20 at 17:27
  • @I'dlikeanonymity even $[0,1]$ doesn't seem trivial. Don't know. – freakish May 28 '20 at 13:08