Before asking a question, please let me write down 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.


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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