On page 63 of Hatcher's book on Algebraic Topology, he says the following:

...one says $X$ is semilocally simply-connected if this holds. To see the necessity of this condition, suppose $p : \widetilde{X} \to X$ is a covering space with $\widetilde{X}$ simply-connected. Every point $x \in X$ has a neighborhood $U$ having a lift $\widetilde{U} \subseteq \widetilde{X}$ projecting homeomorphically to $U$ by $p$.

Is the lift $\widetilde{U}$ just one of the slices of $U$? That is, given $x \in X$, we can find an open neighborhood $U$ of $x$ evenly covered by $p$. I.e., $p^{-1}(U) = \bigcup_{i \in I} U_i$, where $\{U_i\}$ is a disjoint collection of open sets such that $U_i$ is mapped homeomorphically to $U$ by $p$. So $\widetilde{U}$ can be taken to be anyone of the of the sets in $\{U_i\}$? Is this what Hatcher is saying?


I thought I was able to prove his claim that $X$ having a simply connected cover entails that $X$ is semilocally simply-connected, but I was mistaken. Here is what I came up with

Let $\iota_1 : U \to X$ and $\iota_2 : \widetilde{U} \to \widetilde{X}$ denote the canonical embeddings. Let $x \in U$ be an open set in $X$ with lift $\widetilde{U}$. Then $q : \widetilde{U} \to U$ defined as $q := p \big|_{\widetilde{U}} : \widetilde{U} \to U$ is a homeomorphism and hence a covering map. Let $\gamma$ be some loop in $U$ at . Then, because $q$ is a homeomorphism, $\gamma$ tilde lifts to the unique loop $\widetilde{\gamma}^q$ at $\widetilde{x} \in q^{-1}(\{x\})$ (the inverse image is actually just a singleton). Then $\iota_2 \circ \widetilde{\gamma}^q$ is a loop in $\widetilde{X}$ at $\widetilde{x}$. Since $\widetilde{X}$ is simply connected, $\iota_2 \circ \widetilde{\gamma}^q$ must be homotopic to the constant path at $\widetilde{x}$; let $f_t : I \to \widetilde{X}$ denote a homotopy between them. Then $q \circ f_t$ is a homotopy between $q \circ (\iota_2 \circ \widetilde{\gamma}^q)$ and the constant path at $1_x$...

I'm probably just being a knucklehead at the moment, but why does $q \circ (\iota_2 \circ \widetilde{\gamma}^q) = \iota_1 \circ \gamma$? That's what we need in order to conclude that the induced map from $\pi_1(U,x) \to \pi_1(X,x)$ is trivial.

  • $\begingroup$ Yes. Do you know how to proceed from here to show that $X$ must be semilocally simply-connected ? $\endgroup$ – Berni Waterman Mar 6 at 15:38
  • $\begingroup$ @BerniWaterman I thought I was able to figure it out, but I ran into a complication when trying to flesh out the details of Hatcher's "proof". Please see my edit. $\endgroup$ – user193319 Mar 13 at 13:05
  • $\begingroup$ One problem is you need to apply the global projection map $p$ rather than the local one $q$ because the null-homotopy $f_t$ might leave $\tilde{U}$ (so in fact the composition $q\circ f_t$ might not be defined). Another thing that I think adds notational confusion is the $\iota$ maps, because they are just inclusions of subspaces, so let's ignore them. Then $p\circ f_t$ is defined and is a null-homotopy of $p\circ \tilde{\gamma}^q$ (which might leave $U$), and by definition $p\circ \tilde{\gamma}^q = \gamma$. $\endgroup$ – William Mar 13 at 13:25
  • $\begingroup$ I put these details into my answer so hopefully it looks more complete to you now. $\endgroup$ – William Mar 13 at 13:36
  • 1
    $\begingroup$ Yes, the inclusions might be the cause of your confusion. But of course, you can choose them in such a way so that the resulting diagram is commutative, which should resolve all your problems. $\endgroup$ – Berni Waterman Mar 13 at 15:06

Yes, this is what Hatcher is saying. If you consider such an open set $U$ covered by $\tilde{U}$, and an arbitrary loop $\gamma$ in $U$, you should be able to use simple-connectivity of $\tilde{X}$ to deduce semi-local simple-connectivity of $X$.

Details: Let $X$ be a space with basepoint $x_0$. Suppose $p\colon \tilde{X}\to X$ is a simply-connected covering space, and choose a basepoint $y_0\in p^{-1}(x_0)$. Let $U\subset X$ be a neighbourhood of $x$ so that $\tilde{X}$ is trivial over $U$ (like you've suggested), and choose $\tilde{U}$ to be the component of $p^{-1}(U)$ containing $y_0$.

If we choose a loop $\gamma\colon [0,1]\to U\subset X$ based at $x_0$, by local triviality we can lift it for free to a loop $\tilde{\gamma}\colon [0,1] \to \tilde{U}\subset \tilde{X}$ based at $y_0$, i.e. $p\circ \tilde{\gamma} = \gamma$. Since $\tilde{X}$ is simply-connected, there is a null-homotopy $H\colon I\times I \to \tilde{X}$ of $\tilde{\gamma}$ which fixes $y_0$ (and which might leave $\tilde{U}$), and so the projection $p\circ H$ gives a null-homotopy of $p\circ \tilde{\gamma} = \gamma$ through $X$ (which might leave $U$). Since this argument works for all basepoints and all $U$ such that the covering is locally trivial, it follows that $X$ is semi-locally simply-connected.

Aside about the definition: The definition of "semi-local simple-connectivity" has a subtle quirk that might not be immediately apparent. The definition is

$X$ is semi-locally simply-connected if for every $x_0\in X$ there is a neighbourhood $U$ of $x_0$ such that every loop $\gamma$ in $U$ based at $x_0$ is null-homotopic in $X$.

In particular the null-homotopy of $\gamma$ is allowed to leave $U$! Hence you can rephrase by saying that $\pi_1(U,x_0) \to \pi_1(X, x_0)$ needs to be trivial.

There is a stricter condition called "local simple connectivity":

$X$ is locally simply connected if for every $x_0\in X$ and every neighbourhood $U$ of $x_0$ there is a (possibly different) neighbourhood $V\subset U$ such that $\pi_1(V, x_0) = 0$.

For this condition, the null-homotopies DO have to take place in $V$.

Locally simply connected implies semi-locally simply connected, but not the other way around. Spaces which are semi-locally simply connected but not locally simply connected are weird, but they are out there. The standard example seems to be the cone on the Hawaiian Rings: small neighbourhoods of "the bad point" have uncountably many distinct loops which are not null-homotopic in that or any smaller neighbourhood, but since the space is a cone (and hence contractible) every loop is null-homotopic if you're allowed to leave the local area.

  • $\begingroup$ The proof you give of necessity is essentially the one Hatcher gives. However, you (just like Hatcher) leave out some details/subtleties (at least I think so); I tried to flesh them out, but I ran into some complications. Would you mind taking a look at my edit? $\endgroup$ – user193319 Mar 13 at 13:07

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