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Can someone help me understand the difference between locally simply-connected and semilocally simply-connected? I actually need more help understanding (with an example), what it means to be locally simply-connected.

I know that a space X is semi-locally simply connected if every point in X has a neighborhood U for which the homomorphism from the fundamental group of U to the fundamental group of X, induced by the inclusion map of U into X, is trivial. And I understand why the Hawaiian earring is NOT semilocally simply-connected.

Another example that I was trying to understand was that "$CX = (X \times I) / (X \times \{0\})$ (where $X$ is the Hawaiian earring) is semi-locally connected since its contractible, but its not locally simply-connected" - I didn't get why its not locally simply-connected as I don't know how to check for locally simply-connectedness.

edit: My understanding now is that a part of the intermediate loop in the homotopy is permitted to be outside of U, for a semilocally simply-connected space, but cannot be allowed for a locally simply-connected space. But my question about CX remains.

cbro
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    A space $X$ if locally simply connected if for every open neighbourhoud $V$ of each point $x\in X$ there is an open neighbourhood $U\subset V$ of $x$ which is simply connected in the subspace topology of $X$.

    As for $CX$, let $X$ be a not locally simply-connected space and take an open neighbourhood $V$ in $CX$ of $x\in X$ by cutting off the top of the cone. Then since $X$ is not locally simply-connected we won't be able to find the requried $U$. For example let $X$ be the comb space. then $V$ is essentially just $X\times [0,\epsilon)$ and it is easy to see how this fails.

     – Tyrone Jul 20 '17 at 07:49

1 Answers1

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First I would like to point out you say the Hawaiian Earring $X$ is semilocally simply connected when it is not. Any neighborhood of the limit point will contain a full circle (infinitely many but we only need one) and the inclusion will send the generating loop for that circle to the same generator of the fundamental group of $X$ thus not the trivial map.

Now to start, let's state the definition of locally simply connected

A space $X$ is locally simply connected if for every $x \in X$ and every neighborhood $V$ of $x$ there is a neighborhood $U \subset V$ of $x$ that is simply connected in the subspace topology.

So take the cone $CX$ of the Hawaiian Earring and examine neighborhoods of the point $\{x\} \times \{1\}$ where $x$ is the limit of the shrinking circles in $X$. Any small neighborhood $U$ of this point deformation retracts onto a neighborhood of $x$ in $X$ (it looks like $U \times I$ maybe with some of the sides shortened but that doesn't change anything homtopically) and thus $U$ is not simply connected. So we can conclude $CX$ is not locally simply connected.

What makes the difference for semilocally simply connected is you can any neighborhood in $CX$, doesn't matter what the fundamental group is and we have the inclusion $$i:\pi_1(U) \hookrightarrow \pi_1(CX) = 0$$ must be the trivial map.

Birch Bryant
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  • Thanks for the explanation. I meant to say that X is NOT semilocally simply connected actually. So, I will edit my question. – cbro Jul 22 '17 at 00:17
  • Birch Bryant, in "Any small neighborhood $U$ of this point deformation retracts onto a neighborhood of $x$ in $X$ (it looks like $U \times I$ maybe with some of the sides shortened but that doesn't change anything homtopically)", did you mean "Any small neighborhood $U$ of this point deformation retracts onto a neighborhood $V$ of $x$ in $X$ (it looks like $V \times I$ maybe with some of the sides shortened but that doesn't change anything homtopically)"? – João Alves Jr. May 12 '22 at 18:11