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A topological space $X$ is semi-locally simply connected if, for any $x\in X$, there exists an open neighbourhood $U$ of $x$ such that any loop in $U$ is homotopically equivalent to a constant one in $X$ or, equivalently, if the functor $$\Pi_1(U)\rightarrow\Pi_1(X)$$ induced by the inclusion $U\subseteq X$ factorizes through a groupoid in which for each pair of objects there is exactly one morphism.

My question is: is it true that if a space $X$ is such that, for any open subset $U\subseteq X$ ($X$ included), $U$ is semi-locally simply connected, then $X$ must be locally simply connected?

Notice that this implies that $X$ is locally path connected. I tried to post this question on stack exchange some months ago, but I didn't receive any answer. It was just something that came to my mind while I was studying for my master thesis.

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    $\begingroup$ So, are you asking: does "locally semi-locally simply connected" imply "locally simply connected"? If so I love this question. $\endgroup$ – cgodfrey Sep 25 at 4:09
  • $\begingroup$ Yes, it's something like that. If it's needed I might give some more motivation for the question, and maybe give a link to my master thesis, but I don't think that it could really be helpful. $\endgroup$ – mfox Sep 25 at 7:25
  • $\begingroup$ In some sense, I expect that the answer should be positive: a "lower dimensional analog" of this fact is that a space il locally path connected if and only if it is path connected im kleinen at each point. $\endgroup$ – mfox Sep 25 at 7:45
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Here is a counterexample:

Imagine the cone on the Hawaiian earring, but the cone on each circle is half the height of the previous one. Call this space $X$. Any open subset of $X$ is semi-locally simply connected because given a small, connected open neighborhood around the obvious trouble point $x$ at the base of the cone, it must deformation retract to a subspace $V$ homeomorphic to $X \vee S^1 \vee \dots \vee S^1$. This makes it obvious how to choose a neighborhood of $x$ in this subspace $V$ so that every loop in it is nullhomotopic in $V$. Simply take a neighborhood not including any full circle which doesn't have a cone on it. $X$ is not locally simply connected by the exact same observation: there will always be a circle without a cone on it in any small neighborhood of $x$.

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    $\begingroup$ Certainly, there is a neighborhood base at $x$ like you describe. So not all connected neighborhoods of $x$ are simply connected. But it also looks like there is a neighborhood base at $x$ consisting of open sets that deformation retract onto a subspace homeomorphic to $X$, which is simply connected. This would mean $X$ is locally simply connected at $x$. $\endgroup$ – Jeremy Brazas Sep 25 at 14:19
  • $\begingroup$ Ah, I see what you mean @JeremyBrazas. $\endgroup$ – Connor Malin Sep 25 at 14:30
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    $\begingroup$ The construction needs to be changed so instead of $V$ being $X \vee S^1 \vee \dots$ it is $X$ wedge a Hawaiian earring. I suspect it can be done. $\endgroup$ – Connor Malin Sep 25 at 14:36
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    $\begingroup$ But then you lose semilocal simple connectivity... If you cone that off, you run into the same issue. This problem is surprisingly subtle. I doubt a simple combination of the Hawaiian earring and it's cones will provide a counterexample. $\endgroup$ – Jeremy Brazas Sep 25 at 16:56

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