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.

  • $\begingroup$ 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. $\endgroup$ – Zhen Lin Apr 22 '12 at 21:08
  • $\begingroup$ You should give a definition of what you call an action of a groupoid. $\endgroup$ – AFK Apr 22 '12 at 21:11
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    $\begingroup$ @YBL: I am talking about functors from $\Pi_1(X)$ into sets. $\endgroup$ – Justin Campbell Apr 22 '12 at 21:18
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    $\begingroup$ @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 $X$. $\endgroup$ – user8268 Apr 22 '12 at 22:05
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    $\begingroup$ @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) $\endgroup$ – user8268 Apr 23 '12 at 7:32

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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    $\begingroup$ Very nice! I've heard of coverings of groupoids, but I was not aware of result #2. $\endgroup$ – Justin Campbell Apr 23 '12 at 13:27
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    $\begingroup$ 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. $\endgroup$ – Ronnie Brown Apr 23 '12 at 19:45

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