# Lift of an Open Neighborhood

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?

EDIT

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.

• Yes. Do you know how to proceed from here to show that $X$ must be semilocally simply-connected ? – H1ghfiv3 Mar 6 '19 at 15:38
• @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. – user193319 Mar 13 '19 at 13:05
• 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 $p\circ \tilde{\gamma}^q = \gamma$. – William Mar 13 '19 at 13:25
• I put these details into my answer so hopefully it looks more complete to you now. – William Mar 13 '19 at 13:36
• 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. – H1ghfiv3 Mar 13 '19 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.

• 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? – user193319 Mar 13 '19 at 13:07