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One approach to classifying the coverings of a nice space $X$ without choosing a basepoint is to look at actions of the fundamental groupoid on sets. Another way that seems natural to me is to fix a universal covering $p: \widetilde{X} \to X$ and consider the group $\text{Aut}_X(\widetilde{X})$ of deck transformations of $\widetilde{X}$ in place of the fundamental group.

Here's one thing that confuses me about this: if $X$ is connected and $x \in X$, then $\text{Aut}_X(\widetilde{X})$ acts simply transitively on the fiber $p^{-1}(x)$, and in particular we get an isomorphism $\text{Aut}_X(\widetilde{X}) \cong \pi_1(X,x)$. Since this works for any $x,y \in X$, it seems that we get canonical identifications $\pi_1(X,x) \cong \pi_1(X,y)$, but I thought that this could not be done for nonabelian $\pi_1$.

My other question: how useful is $\text{Aut}_X(\widetilde{X})$ when $X$ is nice but disconnected? Do we have some correspondence between actions of this group and coverings of $X$? It is not clear to me what this group looks for, say, $X = S^1 \coprod S^1$.

Edit: I was, of course, incorrect in my formulation of the first question (thanks countinghaus). But the second question stands: do $\text{Aut}_X(\widetilde{X})$-sets correspond to coverings of $X$ even when $X$ is not connected?

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3 Answers 3

For question one: Just knowing that two groups act simply transitively on the same set set isn't enough to give you a canonical isomorphism between them; you still must pick an element of the set to get the isomorphism. This is equivalent to choosing a path between the two basepoints, which is the old-school way to get a (noncanonical) iso on $\pi_1$s with varying basepoints.

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OK, but in this case there is a canonical map: an element of the fundamental group $\pi_1(X,x)$ permutes the fiber $p^{-1}(x)$, which extends uniquely to a deck transformation of $\widetilde{X}$. Right? –  Justin Campbell Apr 24 '12 at 23:38
    
Never mind, I see why this doesn't work. –  Justin Campbell Apr 25 '12 at 0:15

Oops: this was a pretty stupid question. The group $\text{Aut}_X(\widetilde{X})$ just decomposes as $\Pi_i \text{Aut}_{X_i}(\widetilde{X}_i)$ where $X = \coprod_i X_i$ is the decomposition of $X$ into connected components.

I still do not understand, however, what $\text{Aut}_X(Y)$ looks like for an arbitrary (not necessarily connected) cover $Y \to X$, such as a cover $S^1 \coprod S^1 \to S^1$.

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My first guess is that it's a subgroup of a wreath product: you have various isomorphic components of Y, and your automorphism permutes them in some fashion while also acting by automorphisms. So if all the components are isomorphic, I believe you'll get a wreath product of a symmetric group with the automorphisms of one component. With non-isomorphic components, you'll get a direct product of such groups over the various isomorphism types. –  Dan Ramras Apr 26 '12 at 21:34
    
@DanRamras: Very good! I had not heard the term "wreath product" before. This makes it clear to me what's happening. –  Justin Campbell Apr 30 '12 at 17:57

As I've suggested elsewhere on this site, it seems to me that the easier way to look at coverings of $X$ is to use the equivalences of categories

$$\pi_1: TopCov(X) \to GpdCov(\pi_1 X)\to Oper(\pi_1 X)$$

from covering maps to $X$ to covering morphisms to $\pi_1 X$, for suitably nice $X$, and then where the last category is the functor category $\pi_1 X \to Sets$ of operations of the groupoid on Sets. I think it is then quite east to analyse the last two categories if $X$ is not connected. One is asking: if $G= G_1 \sqcup G_2$, how to describe $Oper(G)$?

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