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Briefly, my question is whether there is a basepoint-free statement of the basic theorem on covering spaces. For a nice space $X$, I would hope that there is an equivalence of categories between covering spaces of $X$ and actions of the fundamental groupoid $\Pi_1(X)$ on sets. I can see a functor from covering spaces to $\Pi_1(X)$-sets using the monodromy action of $\Pi_1(X)$ on fibers, but how is the functor in the opposite direction constructed?

Is this the correct setup? I feel like one should be able to carry through all of the constructions with fundamental groups in a basepoint-free way using groupoids.

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    This probably counts as cheating, but couldn't you just take connected components of $X$ and arbitrary base points in each connected component? At least this would give you the functor.2012-04-22
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    You should give a definition of what you call an action of a groupoid.2012-04-22
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    @YBL: I am talking about functors from $\Pi_1(X)$ into sets.2012-04-22
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    @JustinCampbell: Your suggestion is perfectly correct (for locally path connected and semi-locally 1-connected spaces). To go in the opposite direction: you have a set $Y$ and a map $p:Y\to X$ (where $p^{-1}(x)$ is the set corresponding to $x\in X$); the thing is to define a topology on $Y$, which can be done by lifting a basis of the topology on $X$.2012-04-22
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    @user8268: How does one define the topology? It can't be too easy. After all, this theorem implies the existence of a universal covering (apply the construction to the regular action of $\Pi_1(X)$ on itself), which is a nontrivial fact. For this reason I expect the most economical description of the functor to involve a universal covering in some way.2012-04-22
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    @JustinCampbell: for basis on $X$ you can take open $U$'s s.t. $U$ is arc-connected and $\pi_1(U)\to\pi_1(X)$ is trivial (they form a basis by the hypothesis on $X$). Now pick $y\in Y$ s.t. $p(y)\in U$ and get $U'\subset Y$ using the action of $\Pi_1(X)$ (restricted to paths in $U$). These $U'$'s form a basis of topology on $Y$ and $p$ becomes a covering (of course it has to be proved, and it is roughly the same as constructing the universal cover)2012-04-23
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    @user8268: This is quite interesting. So whereas most people define the underlying set of the universal cover to consist of homotopy classes of paths out of some basepoint, we could instead use $\coprod_{x \in X} \pi_1(X,x)$?2012-04-25

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I think the most natural way of doing this is to define first a covering morphism $p: Q \to G$ of groupoids; the definition is essentially unique path lifting: more specifically, we require that if $x \in Ob(Q)$ and $g \in G$ has source $px$, then there is a unique $h \in Q$ with source $x$ such that $p(h)=g$. There are several main results:

  1. A covering map of spaces induces a covering morphism of fundamental groupoids.

  2. The category of covering morphisms of a groupoid $G$ is equivalent to the category of actions of $G$ on sets.

  3. If $X$ is a space, and $q: Q \to \pi_1 X$ is a covering morphism of groupoids, then under certain local conditions on $X$ there is a topology on $Ob(Q)$ such that the map $Ob(q)$ becomes a covering map and ...(I'll leave you to fill in the rest!).

This treatment was in the 1968, 1988 editions of my book which is now available as Topology and Groupoids. The nice point is that a map of spaces is well modelled by a morphism of groupoids; this is convenient for questions on liftings. Also, under the right local conditions, you get an equivalence of categories from covering spaces of $X$ to covering morphisms of $\pi_1 X$.

This is also an introduction to the useful notion of fibration of groupoids.

I think I should mention that the book also has a treatment of orbit spaces and orbit groupoids, which you won't find elsewhere.

Later: A related use of covering morphisms of groupoids is in the paper:

J. Brazas, "Semicoverings: a generalisation of covering space theory", Homology, Homotopy and Applications", 14 (2012) 33-63.

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    Very nice! I've heard of coverings of groupoids, but I was not aware of result #2.2012-04-23
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    I should have added that if $X$ admits a universal cover, then $\pi_1$ gives an equivalence of categories between covering maps to $X$ and covering morphisms to $\pi_1 X$. This is useful also for considering the case $X$ is a not necessarily connected topological group.2012-04-23