**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.