# A technical detail in the construction of a covering space: Is the projection $p$ continuous without assuming semilocal simple connectedness?

I am studying the construction of a covering space with prescribed characteristic subgroup. For simplicity, I will outline the case where the characteristic subgroup is trivial (i.e. we're dealing with the universal cover). Note that the following is very similar, but not identical to the section called "The Classification of Covering Spaces" in Hatcher's book. In particular, Hatcher assumes semilocal simple connectedness from the get-go, while I attended a lecture where the professor explicitly postponed making this assumption as long as possible.

We construct a covering space $E$ of a fixed base space $X$ (locally path connected and connected) in terms of homotopy classes of paths in $X$. The covering projection $p$ sends $[\sigma]$ to $\sigma(1)\in X$. Consider an open and path connected neighborhood $U$ of the end point $\sigma(1)$ of some path $\sigma$. Then, we define an open set in $E$ as

$$U_{[\sigma]}=\{[\sigma*\tau]\mid \tau:(I,0)\to (U,\sigma(1))\text{ a path}\}$$

and claim that the all sets of this form constitute a basis of the topology of $E$ (which can be verified with no problem). Now, we consider the projection $p$. It is clear that $p$ is surjective since local path connectedness plus connnectedness implies path connectedness. $p$ is also an open map (by construction) since $p(U_{[\sigma]})=U$.

Now comes the crucial part (for me): continuity of $p$. In the notes from my lecture we simply find the following statement:

$p$ is continuous because $[\sigma]\in U_{[\sigma]}\subset p^{-1}(U)$ by construction.

This, to me, is definitely not clear. Hence, I tried to piece together a more detailed proof, proceeding along the same lines as Hatcher (last paragraph of page 64-first paragraph of page 65), but as pointed out before one has to take great care in following Hatcher because he assumes semilocal simple connectednes (and hence has e.g. local injectivity of $p$, which I am do not have!). Therefore, I'm having a hard time figuring out how to proceed. Even understanding whether Hatcher's proof of continuity really relies on semilocal simple connectedness has proved to be beyond me, so far.

My question, therefore, is: Is it true that $p$ is continuous, if one does not assume semilocal simple connectedness? If so, an answer would ideally outline the proof in more detail than what I found in my lecture notes (or explain to me why that brief statement suffices). If not, a counterexample would be much appreciated.

• A few comments. (1) The Hatcher paragraph in question does much more than prove continuity of $p$. It also proves that $p$ is a covering map, and that's where semilocal simple connectivity comes into play. (2) Your parenthetical remark on injectivity does not make much sense. Injectivity of $p$ is not at issue in this passage of Hatcher, and that's a good thing because it is generally false. – Lee Mosher Feb 23 '16 at 13:46
• @LeeMosher (1) I realize that that's what he's proving. In my notes, this is not the case because semilocal simple connectedness is not yet assumed. (2) My remark says exactly the same as your comment, I don't understand what the confusion is here. I agree with you that semilocal simple connectedness guarantees injectivity, and that it fails in general (which is why I do not have it "available" in my notes, and Hatcher does). – Danu Feb 23 '16 at 13:52
• In (1), my intent is to pin down more precisely where semilocal simple connectivity is needed: as you ask, it is not needed in proving continuity of $p$; instead, it is used in proving that $p$ is a covering map. My mathematical point in (2) is that semilocal simple connectivity does not imply injectivity of $p$. – Lee Mosher Feb 23 '16 at 13:56
• @LeeMosher I'm sorry, in (2) of course you are correct that $p$ is not injective (globally) but its restriction to appropriate open sets (like $U_{[\sigma]}$) in $E$ is, and this is guaranteed by the assumption of semilocal simple connectedness. Surely, you can agree to that? – Danu Feb 23 '16 at 14:16
• Yes, that sounds correct. – Lee Mosher Feb 23 '16 at 16:25

Yes, $p$ is continuous regardless of whether $X$ is semilocally simply connected or not:

Let $[\sigma] \in E$ so that $p([\sigma]) = \sigma(1) = x_0$ and $U$ be an open neighborhood of $x_0$. By local path connectedness, pick a path connected neighborhood $V$ of $x_0$ contained in $U$. As $p(V_{[\sigma]}) = V$, $V_{[\sigma]}$ is contained in $p^{-1}(V) \subset p^{-1}(U)$, hence is an open neighborhood of $[\sigma] \in p^{-1}(U)$.

Similarly, for any $[\gamma] \in p^{-1}(U)$, I can construct an open neighborhood of $[\gamma]$ contained in $p^{-1}(U)$. Thus, every point in $p^{-1}(U)$ is an interior point, proving $p^{-1}(U)$ is open.

• (Semilocal simply connectedness is assumed to ensure the sets defined are a basis for some topology on $E$. Continuity is ensured by local connectedness.) – Pedro Tamaroff Jul 31 '16 at 3:26

In my book Topology and Groupoids (T&G) I develop the classification of covering spaces using the notion of covering morphism $q: Q \to G$ of groupoids.

This is a morphism of groupoids such that for each $y \in Ob(Q)$ and element $g \in G$ starting at $q(y)$ there is a unique $h \in Q$ starting at $y$ and such that $q(h)=g$. (If one drops "uniqueness" one get the notion of fibration of groupoids.)

If $p: Y \to X$ is a covering map of spaces, one proves by standard methods that $\pi_1(p) : \pi_1(Y ) \to \pi_1(X)$ is a covering morphism of groupoids.

So one is interested in the reverse process: given a covering morphism $q$ as above but with $G=\pi_1(X)$, can we find a $p:Y \to X$ such that such that $q$ is essentially $\pi_1(p)$?

A basic result on covering morphisms one uses here is that if $f: F,z \to G,x$ is a morphism of groupoids which is pointed, i.e. $f(z)=x$, and $F$ is connected, then $f$ lifts to a morphism $\tilde{f}: Z,z \to Q,y$ such that $q\tilde{f}=f$ if and only if $f(F(z) ) \subseteq q(Q(y))$; and if this lift exists it is unique.

This is easy to prove by choosing, using connectivity of $F$, elements $\tau(w): z \to w$ in $F$ for each $w \in Ob(F)$ with $\tau(z)=1_z$. (The detailed proof is in 10.3.3 of T&G.)

Now given $G= \pi_1(X)$, we want to get a topology on $Y=Ob(Q)$. We want to choose a path connected neighbourhood $N$ of $x \in X$ such that if $F=\pi_1(N)$ then the pointed morphism $f: F,x \to G,x$ lifts to a morphism $F,x \to Q,y$ for any $y \in q^{-1}(x)$. Hence we must assume that for all $x \in X$ there is a path connected neighbourhood $N$ of $x$ such that the image of $\pi_1(N,x)$ in $G$ lies in the intersection of the $q(Q(y))$ for all $y \in q^{-1}(x)$. Assuming this, it is not hard to prove that the images of these neighbourhoods in $Y=Ob(Q)$ give a topology on $Y$ making $Q \cong \pi_1(Y)$. The details are in Section 10.5 of T&G.

So to be more specific to the question, without the semilocal condition, one can't construct the topology by this method; with the condition, we can construct the topology on $Y=Ob(Q)$ and prove $p=Ob(q)$ is continuous and a covering map.

The algebraic theory of covering morphisms of groupoids then gives an equivalence of categories of covering morphisms of a groupoid $G$ and functors $G \to Sets$.