Implications between notions of universal covering in categorical Galois theory A while ago I made a generally wrong calculation which led me to ask whether the endomorphisms of a universal cover of a space are necessarily automorphisms.
Going through some papers of Janelidze, remark 5.2 of the paper Categorical Galois theory - revision and some recent developments touches related matters and even affirms my suspicion. However, I am struggling with filling in the details. For the sake of completeness we give some definitions. Throughout, we fix an admissible adjunction $\Pi_0\dashv H$ and suppose the domain of $\Pi_0$ has pullbacks.
Definitions.


*

*An object is Galois closed if its coverings are trivial.

*A covering $\alpha:A\to B$ is connected if $\Pi_0\alpha$ is an isomorphism.


We shall say an arrow splits a covering morphism if pulling back along the arrow trivializes the covering morphism.
Definitions. An effective descent covering morphism $p:E\to B$ is said to be:


*

*Maximal if $E$ is Galois closed.

*Weakly universal if it splits all coverings of $B$.

*Projective if it factors through any other effective descent morphism to $B$.

*Universal if it's weakly universal, projective, and every endomorphism is an automorphism.


Remark 5.2.


*

*Maximal $\implies $ weakly universal. The converse is true if covering morphisms are closed under composition.

*If a weakly universal covering exists, projective $\implies$ weakly universal.

*If a weakly universal covering exists, connected projective $\implies $ universal.


I managed to prove the first implication of (1) and also (2) in a fairly straightforward manner. However I am lost with the converse of (1) and also with (3). I feel I must be missing simple diagrams. For (3) I suspect the orthogonality of arrows inverted by $\Pi_0$ and trivial coverings should be used, but I don't see how to apply it.
 A: For the converse of (1) : let $q:E'\to E$ be a covering, and assume that $pq:E'\to B$ is a covering. Since $p$ is weakly universal, it splits itself and $pq$. So in the diagram
$$\require{AMScd} \begin{CD}E'@>>> E\times_B E' @>>> E'\\
@V{q}VV (1) @V{\pi_2^*(q)}VV (2) @VV{q}V \\ 
E@>{\delta}>> E\times_BE @>{\pi_2}>> E\\
& @V{\pi_1}VV (3) @VV{p}V\\
& & E@>>{p}>B\end{CD}$$
where the squares $(2)$ and $(3)$ are pullbacks, $\pi_1$ and $\pi_1\circ \pi_2^*(q)$ are trivial, and thus $\pi_2^*(q)$ is trivial (as one can see by applying the classical pullback lemma to the naturality squares). Now if the top arrow of $(1)$ is chosen in such a way that the top composition is the identity on $E'$, then the rectangle $(1)+(2)$ is a pullback, and thus so is the square $(1)$. Since trivial extension are stable under pullbacks, $q$ is trivial.
For (3), if $f:E\to E$ be such that $pf=p$. Since $p$ is weakly universal by (2), the same argument shows that $\pi_2^{*}(f)$ is trivial in the sense that
$$\begin{CD}E\times_B E @>{\eta_{E\times_B E}}>> H\Pi_0(E\times_B E) \\ @V{\pi_2^{*}(f)}VV @VV{H\Pi_0(\pi_2^{*}(f))}V \\
E\times_BE @>{\eta_{E\times_B E}}>> H\Pi_0(E\times_B E) \end{CD}$$
is a pullback. Since moreover $\pi_2$ is a trivial covering (because $p$ splits itself), the pullback of the square $(2)$ (with $f$ instead of $q$) is preserved, and thus in
$$\begin{CD} E@>>> E\times_B E @>>> E @>{\eta_{E}}>> H\Pi_0(E)\\ @V{f}VV (1) @V{\pi_2^*(f)}VV (2) @VV{f}V (4) @VV{H\Pi_0(f)}V\\ E @>>{\delta}> 
 E\times_BE @>{\pi_2}>> E @>>> H\Pi_0(E)\\
\end{CD}$$
the rectangle $(2)+(4)$ is a pullback. Since $(1)$ is also a pullback, $(1)+(2)+(4)$ is also a pullback but this is just $(4)$ since the top and bottom composite in $(1)+(2)$ are identities. Now since $p$ is connected, $\Pi(pf)=\Pi(p)$ is an isomorphism, hence so is $\Pi(f)$. Since $(4)$ is a pullback, it follows that $f$ is an isomorphism.
