# Can I test the property to be an epimorphism of sheaves on a sort of 'generalized covers' of representables?

Let $C$ be a Grothendieck-site with finite limits (perhaps I have to assume that the topology it defines is subcanonical, but it would be nice if I wouldn't have to) and let $T=Shv(C)$ denote the category of sheaves on it.

Let $f:F\to G$ be a morphism in $T$ and suppose that for all $X\in T$ there exists objects $Y_{i,X}\in T$ and an epimorphism $\coprod Y_{i,X}\to X$ in $T$ such that the map of sets $$(*)\hspace{3em}Hom_T(Y_{i,X},f):Hom_T(Y_{i,X},F)\to Hom_T(Y_{i,X},G)$$ is surjective. How can I deduce that $f:F\to G$ is an epimorphism of sheaves?

Remark: I am pretty sure, that this is true since if for every representable $X$ the $Y_{i,X}$ would be representable, they would cover $X$ and condition (*) would imply that $f$ is an epimorphism of sheaves. However, I cannot show this when the $Y_{i,X}$ are not representable. I have tried to write them as a colimit of representables $Y_{i,X}=colim_\alpha Y_{i,X}^\alpha$ and use $Hom_T(Y_{i,X},f)\cong \lim_\alpha Hom_T(Y^\alpha_{i,X},f)$ but I don't get further.

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## 1 Answer

Well, if the condition holds for all $X$, then in particular it holds for $X = G$. So choose $q_i : Y_i \to G$ such that $q : \coprod_i Y_i \to G$ is an epimorphism and $(*)$ holds. Then there must exist $p_i : Y_i \to F$ such that $q_i = f \circ p_i$, by hypothesis, and so if $p : \coprod_i Y_i \to F$ is the induced morphism, then $q = f \circ p$. But $q$ is an epimorphism, so $f$ must be as well.

Note that the above argument does not depend on the fact that we are working in a topos. Rather, it is the converse that requires something extra!

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Ah, yes, thanks, I was a little blind. Unfortunately, I can't up vote your answer. – sopot Dec 2 '13 at 20:35