# Another Question in Hatcher

First of all, I apologize for asking yet another question about the hypotheses of a problem in Hatcher, but the statement of one of his problems has stumped me again.

The problem is 1.3.15. It reads as follows:

Let $p:\widetilde{X}\rightarrow X$ be a simply-connected covering space of $X$ and let $A\subseteq X$ be a path-connected, locally path-connected subspace, with $\widetilde{A}$ a path-component of $p^{-1}(A)$. Show that $p:\widetilde{A}\rightarrow A$ is the covering space corresponding to the kernel of the map $\pi _1(A)\rightarrow \pi _1(X)$.

My question is: why do we need to assume that $A$ is path-connected and locally path-connected?

Here is a sketch of the "proof" I have that does not make use of these hypotheses:

If we have a loop in $\widetilde{A}$, this loop is nulhomotopic in $\widetilde{X}$, and hence will be nulhomotpic when mapped down to $X$ via $p$. On the other hand, if you go the other way and first map the loop in $\widetilde{A}$ down to $A$ via $p$, and then include this loop into $X$, by commutativity of the corresponding diagram I can't draw here, you get the same loop as the first way, which is a nulhomtopic loop. This shows that $\pi _1[p]\left( \pi _1\left[ \widetilde{A}\right] \right)$ is contained in the kernel of the inclusion. Convserly, if we have a loop in $A$ that is nulhomotpic in $X$, we can lift this to a loop in $\widetilde{X}$, and hence we can choose such a lift that is contained in $A'$. Pushing this lift down via $p$, shows that the containment is actually equality.

Any idea where this proof goes wrong?

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If $A$ is not path-connected, then a single path-component of $p^{-1}(A)$ will not even surject onto $A$, so it could not be a covering space of $A$. You seem to have the correct intuition for why the proposition is true, but your argument is implicitly using the hypotheses.

Rather than trying to see why your proof does not work (which can be difficult because it is easy to implicitly incorporate hypotheses without realizing it), it might be more productive to think of examples which do not satisfy the hypotheses (like any $A$ which is not path-connected) and see why the claim fails in this case. Then armed with this counterexample, you can examine your proof to see where the hypothesis was used.

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Having taken your advice, I am no longer worrying about the path-connectedness assumption. However, I am still having difficulty coming up with a counterexample if $A$ is not locally path-connected, mostly because I'm having difficulty thinking of spaces that are path-connected, but not locally path-connected period (only come up with one such space). Any ideas? – Jonathan Gleason Oct 27 '11 at 21:03
Hatcher p.62 (emphasis mine): "Thus for $Y$ to be locally path-connected means that for each point $y \in Y$ and each neighborhood $U$ of $y$ there is an open neighborhood $V \subset U$ of $y$ that is path-connected. Some authors weaken the requirement that $V$ be path connected to the condition that any two points in $V$ be joinable by a path in $U$. This broader definition would work just as well for our purposes, necessitating only small adjustments in the proofs, but for simplicity we shall use the more restrictive definition." PS: Hatcher's covering maps need not be surjective (p.56). – leslie townes Oct 28 '11 at 0:53

Tom's answer is good. I am adding a second one, for the reason that some textbooks do not require covering maps to be surjective. (Added one day later: indeed, Hatcher does not require this.)

The first observation that came to my mind is that if these spaces are not path connected it is bad to notationally suppress the choice of basepoint (e.g. to write $\pi_1(A)$ and not $\pi_1(A, a_0)$ for some fixed and explicitly chosen $a_0 \in A$). This quibble would apply equally well to $X$ itself. So let us bring basepoints to the forefront to make things clearer.

I think $X$ is implicitly being assumed path-connected and locally path-connected (or at least assumed something--- it is not just a random topological space), for the reason that some hypothesis is necessary in order to know that $X$ has a simply-connected covering space in the first place. I don't have Hatcher nearby but I think the construction he gives for $\widetilde{X}$ does have hypotheses like this on $X$.

So far this is just a notational quibble. If you assume basepoints have been chosen in the background so that everything is well-defined (as is often done in situations like this) there is still the issue of whether the hypothesis on $A$ is necessary. With the caveat that this is just based on a casual reading, not hours of thought, my feeling is that probably the hypothesis on $A$ is not necessary, but that there is nothing gained by considering more general $A$.

Let me make my suspicion more explicit.

Suppose $A$ is not path connected. You've chosen some basepoint $a_0$ in $A$; let $A'$ denote the path component of $A$ containing $a_0$. By hypothesis, there is stuff in $A$ that is not in $A'$, but none of this stuff is going to affect "$\pi_1(A)$" (which is of course $\pi_1(A,a_0) = \pi_1(A',a_0)$) at all.

Similarly, when you consider the path-component $\widetilde{A}$ of $p^{-1}(A)$, I think you are implicitly required to choose a path-component of $\widetilde{A}$ that contains an element of $p^{-1} \{a_0\}$. (This goes back to the basic definition of the correspondence between covering spaces of a space $Y$ and subgroups of the fundamental group of a space $Y$. If $f: C \to Y$ is any function at all (in particular if it is a covering map), one must choose basepoints $y_0 \in Y$ and $c_0 \in C$ with $f(c_0) = y_0$ before one can even talk about a homomorphism $\pi_1(C,c_0) \to \pi_1(Y, y_0)$ induced by $f$. So to even talk about the subgroup of $\pi_1(A, a_0)$ of $A$ corresponding to $p: \widetilde{A} \to A$ we must have chosen a point $\tilde{a_0}$ in $\widetilde{A}$ that gets mapped to $a_0$.) With that in mind, since $\widetilde{A}$ is path connected and contains something in $p^{-1}\{a_0\}$, we see that $\widetilde{A}$ actually has to be a subset of $p^{-1}(A')$. It can't contain things that project down to stuff in $A \setminus A'$ (or we could use path connectedness to draw a path in $\widetilde{A}$ from such a thing to $\tilde{a_0}$, and compose this path with $p$ to get a path from $a_0$ to something that isn't in $A'$).

So all of the data in the statement ends up depending only on the component $A'$ in which the basepoint is chosen, and not on the rest of $A$. I think that is why you never need to use the hypothesis explicitly: all of the loops you discuss when you sketch that argument (suppressing the choice of basepoint) are only taking place in one path component of $A$ (whatever part of it your secretly chosen basepoint lies in). It only becomes mysterious when you suppress the choice of basepoint. Anyway, that is my guess about what is going on.

I guess I have only explained why "path connected" is part of the hypotheses--- not the "locally path-connected" piece. Frankly I can never remember why things like that are added in except that you see it all over the place in discussions of fundamental groups, probably because it rules out pathologies. (Revised on edit: see the comment to Tom's answer. I think Hatcher is using it just because it simplifies some proofs. Probably it is not strictly necessary, but if he were to drop it in the exercises, he would have to give more complicated proofs of the results he does prove in the text.)

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I have just come across this question. It does seem a question where the approach via covering morphisms of groupoids, as explained in the 1968, 1988, and 2006 editions of the book now titled and available as "Topology and Groupoids", has clear advantages, since a covering map of spaces induces a covering morphism of groupoids. (This notion occurred under a different name in a 1951 Annals paper of P.A. Smith, and was independently used by Philip Higgins, see his book Categories and groupoids (downloadable). See also Peter May's book "A concise course in algebraic topology", (1999). It may be conveniently generalised to the notion of fibration of groupoids.)

The pullback of a covering map by a continuous map is also a covering map. The pullback of a covering morphism of groupoids by a morphism of groupoids is also a covering morphism. Such a pullback square can be analysed in detail by a Mayer-Vietoris type sequence, involving object groups and components. (This analysis was not in the 1968 edition of T&G.)

Massey's book on "Algebraic Topology: an Introduction" also discusses a pullback of covering spaces, if my memory is correct.

Edit: 15 Feb, 2014 A paper on the Mayer-Vietoris sequence, available here, is

[40]. (R. Brown, P.R. HEATH and H. KAMPS), Groupoids and the Mayer-Vietoris sequence'', J. Pure Appl. Alg. 30 (1983) 109-129.

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