I am confused by the statements of a couple of the exercises in Section 1.3 of Hatcher. I think they need additional hypotheses that are not reflected in Hatcher's errata.

Exercise 11: Construct finite graphs $X_1$ and $X_2$ having a common finite-sheeted covering space $\tilde X_1= \tilde X_2$, but such that there is no space having both $X_1$ and $X_2$ as covering spaces.

Hatcher does not require that the covering map $p: \tilde X\rightarrow X$ be a surjection in his definition on p. 56. So, it seems that for any pair of spaces $X,Y$ they cover a common space $X\sqcup Y$. So it seems that perhaps adding the hypothesis that the covering map be a surjection is the thing to do.

Exercise 31: Show that the normal covering spaces of $S^1\vee S^1$ are precisely the graphs that are the Cayley graphs of groups with two generators.

It seems like we should require that the covering space be connected as well. Because for instance, we may take the cover $\tilde X$ of $S^1\vee S^1$ by two copies of $S^1\vee S^1$ projecting down. There is then a deck transformation swapping these two copies so it is a normal covering space. However, it is not a Cayley graph because $\tilde X$ is not connected.

As a final question, I am curious to know

What are the advantages of not requiring that the covering map be surjective?

In Chapter 1 of Hatcher, we quickly restrict to path connected covering spaces, where the covering map is surjective. So, I have not seen any examples illustrating why this is useful. Perhaps the reason is more advanced?

  • $\begingroup$ Most authors require surjectivity in the very definition. I've never seen an advantage to not doing this. $\endgroup$ – Hoot Jun 28 '16 at 14:53

To your first/last question regarding the surjectivity of the covering map $p : \tilde X \to X$, you're right; in fact Hatcher explicitly states that a covering map need not be surjective. However, Hatcher also explains that the cardinality of the preimage $p^{-1}(x)$ for a given $x \in X$ is locally constant, i.e. this is the number of disjoint sheets in the preimage $p^{-1}(U)$ for a neighborhood $U$ of $x$. In particular, the definition of a covering space $p : \tilde X \to X$ does require that $p$ is surjective if $X$ is connected. Your example $X \sqcup Y$ is a valid example of a nonsurjective covering space, but can you think of why a connected space $W$ cannot be covered by both $X$ and $Y$ for appropriate choices of $X$ and $Y$? (Allen Hatcher pointed this user in the right direction.)

So I guess the answer to your last question is, perhaps it's not so much that there's an advantage to require or to not require that $p : \tilde X \to X$ is surjective. It's that there's an advantage to assuming that $X$ is connected (if it's not then you can just consider how the individual connected components of $X$ are covered by the different connected components of $\tilde X$). And if $X$ is connected, as discussed, $p : \tilde X \to X$ must be surjective.

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