Suppose we're dealing with nice spaces which are the coproducts of their connected components. Then there's a functor $\Pi_0:\mathsf{Top}\rightarrow \mathsf{Set}$ taking a space to its set of connected components.

Let $p:E\to B$ be a covering map. The Galois "groupoid" of $p$ $$\Pi_0(E\times _B E\times _B E)\rightarrow \Pi_0(E\times _BE) \rightrightarrows \Pi_0(E)$$ is defined as the image along $\Pi _0$ of $$E\times _B E\times _B E\rightarrow E\times _BE \rightrightarrows E$$ where the left arrow forgets the middle element of a triple $(x,y,z)$, and the arrows on the right are the pullback projections (the unit arrow is given by the diagonal).

Theorem 6.7.4 of Borceux and Janelidze's Galois Theories says that if $p$ is a universal covering map with connected $E$ (and therefore $B$), then the Galois groupoid is a group isomorphic to $\mathsf{Aut}(p)$ - the group of automorphisms of $p$ (over $B$).

I tried to calculate the Galois "groupoid" of a covering map with connected $E$ (and therefore $B$) without any further assumptions and found it isomorphic to $\mathsf{End}(p)$:

The connectedness of $E$ is equivalent to $\Pi_0(E)=\bf 1$ so there's only one object and the "groupoid" is indeed a group has a single object. To calculate $\Pi_0(E\times _BE)$ we use the distributivity of the category of spaces, the fact covering maps have discrete fibers, and the fact $\Pi_0$ is left adjoint to discrete spaces. This shows $$\Pi_0(E\times _BE)\cong \Pi_0(\coprod_b(p^{-1}\left\{ b\right\}\times p^{-1}\left\{ b\right\} ))\cong \coprod_b \Pi_0(p^{-1}\left\{ b\right\}\times p^{-1}\left\{ b\right\}).$$

Now suppose the fibers are of size $n$. Then $|\Pi_0(p^{-1}\left\{ b\right\}\times p^{-1}\left\{ b\right\})|=n^2$, and it seems like $\Pi_0(E\times _BE)$ is comprised of endomorphisms of $p$.

Is this reasoning correct? Does it imply that endomorphisms of universal covering maps are automorphisms? What's the intuition for this?

Added. My original formulation was misleading: the Galois "groupoid" is not generally a groupoid unless $p$ is a principal bundle. I think it's just an internal category in general. (This sits well with the fact $\Pi_0(E\times _BE)$ is not a group for a general covering map.)

  • $\begingroup$ I don't think $E\times_B E$ is $\sqcup_b p^{-1}\{b\}\times p^{-1}\{b\})$. Take for example $\mathbb R\to S^1=[0,1]/_{0\sim1}$ given by $r\mapsto r-\lfloor r\rfloor$; then $E\times_B E$ is $\{(r_1,r_2)\in\mathbb R^2: r_1-r_2\in\mathbb Z\}$, which is the disjoint union of countably many lines indexed by $\mathbb{Z}$, and not a disjoint union indexed by the uncountably many $b\in[0,1]/_{0\sim1}$. $\endgroup$ Jan 2, 2017 at 2:33
  • $\begingroup$ Your addendum is wrong, $\Pi_0(E\times_B E)$ is always a groupoid, not merely an internal category. It's just that it's not always a Galois groupoid. $\endgroup$ Jan 19, 2017 at 22:12

2 Answers 2


Your definition of the Galois groupoid is incomplete -- it is supposed to be an internal groupoid not an internal category, cf. Example 4.6.2 and Lemma 5.1.22.

Explicitly, in any category $\mathcal C$, the double and triple pullback of a morphism $E\xrightarrow{p}B$ naturally equips $E$ with the structure of an internal groupoid as follows.

  1. The internal domain and codomain maps are the natural projections $E_{p_1}\times E_{p_2}\underset{\pi_2}{\overset{\pi_1}\rightrightarrows} E$. Here $p_1,p_2$ are both $E\xrightarrow{p}B$, and the natural projections are the ones universally satisfying $\pi_1\circ p_1=\pi_2\circ p_2$. The identity map which is required to split both the domain and codomain is the diagonal $E\xrightarrow{\Delta}E_{p_1}\times E_{p_2}$ uniquely satisfying $\pi_1\circ\Delta=\mathrm{id}=\pi_2\circ\Delta$.
  2. The partially-defined composition map has domain the pullback of $E_{p_1}\times E_{p_2}\xrightarrow{\pi_2}E$ along $E_{p_3}\times E_{p_4}\xrightarrow{\pi_3}E$ (meaning: combines a pair of morphisms so that the codomain of the first matches the domain of the second). The composition is given by $(E_{p_1}\times E_{p_2})_{\pi_2}\times(E_{p_3}\times E_{p_4})_{\pi_3}\xrightarrow{(\pi_1\circ\epsilon_1,\pi_4\circ\epsilon_2)}E_p\times E_p$ where $(E_{p_1}\times E_{p_2})_{\pi_2}\times(E_{p_3}\times E_{p_4})_{\pi_3}\underset{\epsilon_2}{\overset{\epsilon_1}\rightrightarrows} E_p\times E_p$ are the projection morphisms of the domain of the composition map. In other words, the composition map is the unique one that combines the two morphisms into a morphism by eliminating their matching codomain and domain, and keeping the resulting domain and codomain.
  3. There is also an inversion map which inverts morphisms. It is given by $E_{p_1}\times E_{p_2}\xrightarrow{(\pi_2,\pi_1)}E_{p_1}\times E_{p_2}$, i.e. it is the unique morphism that takes a morphism and spits out a morphism with switched domain and codomain.

If you are so inclined, you can check by hand that all the various morphisms satisfy the equations of an internal groupoid, or more systematically you can relate iterated pullbacks to a limit-preserving functor from the categories of finite subsets of $\{0,1,\dots,\}$.

The Galois groupoid is the image of the internal groupoid so described under the functor $\Pi_0$ for $E\xrightarrow{p}B$ a morphism of relative Galois descent for the functor $\Pi_0$. Being descent means that $E\xrightarrow{p}B$ makes the pullback functor of a category of morphisms over $B$ monadic over a category of morphisms over $E$ (so it identifies morphisms over $B$ with well-structured morphisms over $E$).

Being relative Galois descent additionally requires that certain morphisms to $B$ are split relative to $\Pi_0$ by $E\xrightarrow{p}B$, the morphisms and the notion of splitting arising naturally from the existence of a right adjoint to $\Pi_0$.

The point then is that $\Pi_0(E\times_B E)$ always has the structure of an internal groupoid on objects $\Pi_0(E)$. However, it is only in the case when the morphism is of relative Galois descent that this groupoid is a well-behaved in the sense that there is an equivalence of categories between internal presheaves for this groupoid and all morphisms to $B$ split relative to $\Pi_0$ by $E\xrightarrow{p}B$; in the case where the groupoid is a group the equivalence is between $G$-sets and such morphisms, a special case of $G$-sets being normal subgroups.

  • 2
    $\begingroup$ Why does $\Pi_0(E\times_B E)$ always have the structure of an internal groupoid? I don't understand why $\Pi_0$ should preserve the required pullbacks in general. The proof of Lemma 5.1.22 relies on lemma 5.1.18 which in turn relies on lemma 5.1.15 which finally relies on $\sigma$ being of Galois descent. $\endgroup$
    – Arrow
    Nov 29, 2017 at 18:18
  • $\begingroup$ I think you're right; I don't know what I was thinking when I wrote that (did I forget that internal groupoids involve pullbacks???). $\endgroup$ Dec 9, 2017 at 23:23

A weakly initial covering morphism of $B$ is a weakly initial object in the category of effective descent covering morphisms of $B$.

A weakly universal covering morphism of $B$ is a covering morphism which trivializes all other covering morphisms of $B$.

A covering morphism is connected if it is inverted by $\Pi_0$.

Theorem. If $B$ admits a weakly universal covering, then any endomorphism of a connected weakly initial covering morphisms is an automorphism.


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