# Space with semi-locally simply connected open subsets

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.

• So, are you asking: does "locally semi-locally simply connected" imply "locally simply connected"? If so I love this question. – cgodfrey Sep 25 '19 at 4:09
• 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. – mfox Sep 25 '19 at 7:25
• 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. – mfox Sep 25 '19 at 7:45

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 $$x$$.
• 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 $x$. – Jeremy Brazas Sep 25 '19 at 14:19
• 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. – Connor Malin Sep 25 '19 at 14:36