Requiring covering spaces of a well-behaved connected topological space $X$ to be connected, let $\mathcal{Cov}(X)$ be the category of covering spaces of $X$ and maps over $X$ and maps over $X$. Can anyone provide me with a quick way to see that if $G$ is the fundamental group of $X$, then $\mathscr{O}_G$ is equivalent, not isomorphic, to $\mathcal{Cov}(X)$? Thanks.
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$\begingroup$ These notes by Jesper Møller should be helpful: arxiv.org/abs/1106.5650 $\endgroup$– Dan RamrasCommented Sep 4, 2015 at 19:41
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1$\begingroup$ MSE is a good forum for this question. $\endgroup$– Ryan BudneyCommented Sep 4, 2015 at 20:30
1 Answer
My book Topology and Groupoids deals with covering maps of spaces by modelling them in terms of covering morphisms of groupoids, an approach which loosens the dependence on base points of the traditional approach. The account is developed from the1968, differently titled, edition of that book, and is partially adopted in Peter May's "Concise..." text.
I learnt of this notion for groupoids from Philip Higgins in the 1960s, and it is developed also in his 1971 notes Categories and Groupoids, now downloadable, with applications to group theory.
I am not sure there is a quick way, but I feel the groupoid approach has advantages: given a topological space $X$ it asks for conditions for a covering morphism of groupoids $q: G \to \pi_1(X)$ to be realised up to isomorphism by a covering map $p: Y \to X$ of spaces, with $G \cong \pi_1(Y)$.
Using the fundamental groupoid also has advantages as shown in this paper: Covering groups of non-connected topological groups revisited, Math. Proc. Camb. Phil. Soc, 115 (1994) 97-110. It brings out the relation of this question to crossed modules.