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Definition. An arrow $\alpha:A\rightarrow B$ in $\mathsf C=\mathsf{Fam}(\mathsf A)$ is said to be a covering morphism if there exists an effective descent morphism $p:E\rightarrow B$ that splits it, i.e such that the square below is a pullback. $$\require{AMScd} \begin{CD} E\times_BA @>{\eta_{E\times_BA}}>> H\Pi_0(E\times_BA)\\ @V{p^\ast\alpha}VV @VV{H\Pi_0(p^\ast\alpha)}V\\ E @>>{\eta_E}> H\Pi_0(E) \end{CD}$$ Here $\Pi_0$ is the connected component functor while $H$ is its right adjoint, defined by taking copowers of $\mathbf 1$.

Definition. An object $E$ is Galois closed if every covering morphism $E^\prime \rightarrow E$ is split by $1_E$.

Definition. A covering morphism $p:E\rightarrow B$ is said to be a universal covering of $B$ if it's an effective descent morphism and $E$ is Galois closed.

For a universal covering morphism $p:E\rightarrow B$, how to prove $E$ connected implies $B$ connected?

For a trivial covering morphism in the concrete setting of topological spaces, I know being of effective descent implies surjectivity. Since for trivial covering morphisms, the connected components of the total space are duplicates of connected components of the base (with multiplicity given by the size of the fiber), it's clear that in the surjective case the total space has at least as many connected components as the base space and that finishes the proof.

However, for the nontrivial case, and more generally, for the nontrivial case in a general setting, I'm clueless. Help!

Added. My question arises from two statements in Borceux and Janelidze's Galois Theories which together confuse me. On one hand, at the top of p.213, the authors write

... is defined only for connected spaces $B$ which admit a universal covering map $p:E\rightarrow B$ with connected $E$...

On the other hand, the beginning of theorem 6.7.4 reads:

Let $p:E\rightarrow B$ be a universal covering morphism with connected $E$ (and therefore also connected $B$).

The latter makes it seem that the connectedness of $B$ is redundant in the former, and, of course - I don't know how to prove it.


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  • $\begingroup$ In topology, the image of a connected space by a continuous map is connected. $\endgroup$ – Fernando Muro Jun 16 '16 at 20:54
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    $\begingroup$ @FernandoMuro yes, but the analogue for categories of the for $\mathsf{Fam}(\mathsf A)$ is merely that if $E$ is connected then $\mathsf{Hom}(E,-)$ preserves coproducts, i.e the arrow $p:E\rightarrow B$ corresponds to an arrow $p:E\rightarrow B_i$ for $B_i$ some connected component of $B$. How to deduce $B$ must be connected from this? $\endgroup$ – Arrow Jun 16 '16 at 20:56
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Here is a general claim: in an extensive category $\mathbf{C}$, if $p: E \to B$ is an epimorphism and $E$ is connected (i.e., $\hom(E, -): \mathbf{C} \to \text{Set}$ preserves finite coproducts), then also $B$ is connected.

Lemma: If a coproduct inclusion $i_U: U \to U + V$ is epic, then $V$ is initial. Proof: let $i_1, i_2: V \rightrightarrows V + V$ be the two coproduct inclusions. Then $i_U$ equalizes $U + i_1, U + i_2: U + V \rightrightarrows U + V + V$. Hence $U + i_1 = U + i_2$ since $i_U$ is epic. Pulling back these arrows along the coproduct inclusion $V + V \to U + (V + V)$, we deduce $i_1 = i_2$ since we are in an extensive category. By disjointness of coproduct inclusions, we find that $V$ is initial.

Proof of claim: It suffices to show that in any coproduct decomposition $B = U + V$, exactly one of $U, V$ is not initial. By connectedness of $E$, the map $p: E \to U + V$ factors through one of the summands, say the inclusion $i_U: U \to U + V$. But since $p$ is epic, so must be $i_U$. This forces $V$ to be initial by the lemma. And then $U$ must not be initial, else by strictness of initial objects, $E$ would be too (and initial objects are never connected).

More details at the nLab.

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  • $\begingroup$ This is elegant! Is my answer correct as well? $\endgroup$ – Arrow Jun 18 '16 at 13:10
  • $\begingroup$ @Arrow Looks good to me! $\endgroup$ – Todd Trimble Jun 18 '16 at 15:22
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A groupoid is connected if and only if the diagram below is a coequalizer, since it means all objects are connected by arrows.

$$\mathsf{Arr}(\mathsf G)\rightrightarrows \mathsf{Ob}(\mathsf G)\rightarrow \mathbf 1$$

$B$ is connected if and only iff $\Pi_0B=\bf 1$. Hence, as the authors write in the proof of Lemma 6.7.1, $\mathsf{Gal}[p]$ is connected if and only if the diagram below is a coequalizer.

$$\Pi_0(E\times _BE)\rightrightarrows \Pi_0(E)\rightarrow \Pi_0(B)$$

If $E$ is connected, the groupoid is a group and in particular connected, so

$$\Pi_0(E\times _BE)\rightrightarrows \Pi_0(E)\rightarrow \bf1$$ is a coequalizer.

Now, $E\times _BE\rightrightarrows E\rightarrow B$ is a coequalizer diagram because $p$ is of effective descent and $\Pi_0$ preserves coequalizers as a left adjoint, so $\Pi_0(B)\cong \bf 1$, i.e $B$ is connected, as desired.

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