**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!