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Before asking a question, please let me write down settings.

SETTINGS:

Let $C$ be a category with fiber products and $B$ be a closed subcategory of $C$ (i.e. $B$ contains any isomorphism of $C$, and any base change of any morphism in $B$ is also a morphism in $B$).

Let $T$ be the topology on $C$ associated to $B$ (i.e. $Cov T$ consists of universal effective epimorphic families {$f_{i}:U_{i}\to U$} in $C$ such that each $f_{i}$ is a morphism in $B$).

Moreover, assume that the topology $T$ satisfies the following two conditions:

Condition I

Let $f,g,h$ be arbitrary morphisms in $C$, and assume that $h=gf$ and $h$ is in $B$.

  1. If $g \in B$, then $f \in B$.

  2. If ${f} \in CovT$, then $g \in B$.

Condition II

For any family of maps {$f_{i}:U_{i}\to X$} in $C$ for which there exists "disjoint union" $\coprod_{i} U_{i}$, then the induced map $\coprod_{i}U_{i}\to X$ is in $B$ if and only if $f_{i} \in B$ for all $i$.

Under the situation above,

Let $R$ be a categorical equivalence relation on an object $U \in C$ such that the two canonical projections $\pi_{i}: R\to U$ ($i=1,2$) are both covering maps of $U$ in the topology $T$.

And assume that this categorical equivalence has $T$-quotient $X$. This means that there exists a categorical quotient $p:U\to X$ of $\pi_{i}:R\to U$ such that the induced morphism of associated sheaves (on $T$) $p_{\ast}:h_{U} \to h_{X}$ is a categorical quotient of $\pi_{i \ast}:h_{R}\to h_{U}$ in the category of sheaves of sets on $T$.

(Then, one can prove that $R \cong U\times_{X}U$.)

Now, this is my QUESTION:

Is the map $p:U\to X$ a covering map in $T$ ?

In other words, is $p$ a universal effective epimorphism and satisfying $p\in B$ ?

I could prove a "converse" statement i.e. if $p:U\to X$ is a covering map of the topology $T$, then $\pi_{i}: U\times_{X}U \to U$ ($i=1,2$) is a categorical equivalence relation such that each $\pi_{i} \in CovT$ and has $p$ as a $T$-quotient.

But I could not prove the statement in my question.

Please give me any advice.

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1 Answer

If by "topology" you mean the usual notion of "Grothendieck topology" (and not something weaker like "Grothendieck pretopology"), then the answer is yes. In fact, if $C$ is any site and $p\colon U\to X$ any morphism in $C$ which becomes an epimorphism in $Sh(C)$, then $p$ is a covering morphism for the topology of $C$.

The reason is that an epimorphism of sheaves $q\colon F\to G$ is precisely one which is "locally surjective": for any $A\in C$ and section $s\in G(A)$, there exists a covering family $c_i\colon B_i \to A$ and sections $r_i \in F(B_i)$ such that $q(r_i) = c_i^*(s)$. Applying this to the representable sheaves on $U$ and $X$, with $A=X$ and $s = 1_X$, we find that if $p$ becomes an epimorphism of sheaves, there exists a covering family of $X$ each element of which factors through $U$. By the saturation condition on a Grothendieck topology, this implies that $p$ is a covering map.

This saturation condition is omitted in the notion of "Grothendieck pretopology", though, so in that case I don't know the answer. (I also suspect that by "topology" you mean "pretopology", since your conditions on $B$ are insufficient to ensure that your definition of $T$ is saturated. So maybe this is not a very good answer.)

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