# When does the left-adjoint to a geometric morphism preserve epis?

Suppose I have a functor $f:(C,J)\to(D,K)$ between Grothendieck sites. Is there a condition on $f$ such that $f_!$ (the left adjoint to $f^*$) sends "$J$-epimorphisms", to $K$-epimorphisms, where by $J$-epimorphism I mean:

$h:X\to Y$ such that for all $C$, and all $y \in Y(C)$, there exists a cover $(g_i:C_i\to C)$ in $J$ and $y_i \in X(C_i)$ such that for all $i$, $Y(g_i)(y)=h(y_i)$.

EDIT: If X and Y are sheaves, then the notion of "J-epimorphism" coinincides with the categorical epis. As mentioned by David Brown, ANY left adjoint will preserves epis.

In fact, in the situation in which I was interested, I actually have such a (appropriate analogue of a) J-epimorphism between a sheaf and a stack, so, since f_! is a left adjoint, it will preserve this.

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(Note that you can use latex as you would normally. It makes subscripty expressions easier to read.) –  François G. Dorais Mar 30 '10 at 18:52
Certainly $f^*$ does not always have a left adjoint (e.g. $f^*$ might not even be exact). If the functor f induces a morphism of sites (so that in particular $f^*$ is exact), I believe $f^*$ still may not have a left adjoint. –  David Zureick-Brown Mar 30 '10 at 19:00
f^* certainly DOES have a left adjoint SINCE the geometric morphism arises from a functor from C to D. –  David Carchedi Mar 30 '10 at 19:10
(the left adjoint is given by the left-kan extension of $C \mapsto Hom( blank, f(U))$) –  David Carchedi Mar 30 '10 at 19:11
@David C: The trick to get good answers is to ask good questions. Well motivated questions usually get excellent answers. As is, your question is hard to understand and motivate. It's only after reading your comment to David B's answer that I understood what was going on. That comment would fit in very well within the question and make it much easier to understand. (A few notational fixes wouldn't hurt either.) –  François G. Dorais Mar 31 '10 at 12:46

A left adjoint functor always takes epimorphisms to epimorphisms. This is easy to see using the usual definition of epimorphism and checking that this is equivalent to your definition for sheaves on a site.

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Ah, well, I actually meant f^* and f_* to be the geometric morphism between the topoi of presheaves, not sheaves, which is why I didn't simply write epimorphism, but decorated it with a "J". However, I realized that in the example I cared about, both of my guys are sheaves (well actually one is a sheaf and one is a stack). So, this is still quite helpful, so you'll get an upvote :-). Anyhow, I'm still interested in the question for presheaves, out of mere curiosity. –  David Carchedi Mar 30 '10 at 19:22
Maybe this works in general: it seems like your definition of `J-epimorphism' is equivalent to the sheafification of the morphism being an epi-morphism in the category of sheaves. Then your question follows from the case of sheaves (since, in any case, you have to sheafify $f_!$ to get a left adjoint to $f^*$). –  David Zureick-Brown Mar 30 '10 at 20:06
By the way David, a left adjoint DOESN'T always take epis to epis! Counter-example: The forgetful functor from Top to Set. (Epis in Top are continous maps with dense image). What goes wrong is that Top is not a regular category! What you're thinking is that left-adjoints preserve REGULAR epis, and this is just because they are coequalizers! BUT, we are saved because in Set, and in Groupoids, all epis are regular. –  David Carchedi Apr 1 '10 at 17:07
@David: "Epis in Top are continous maps with dense image" -- you need Hausdorff for this (just consider the topological space of 3 points with one point open. Then the inclusion is not epic, but it has dense image). For Hausdorff topological spaces there is no right adjoint to the forgetful functor. –  David Zureick-Brown Apr 2 '10 at 0:55
Also, here's the proof that a left adjoint functor L (with right adjoint R) sends epis to epis: let $X \to Y$ be epic and let $L(X) \to L(Y) \to \to Z$ be two maps from L(Y) to Z such that the compositions from L(X) to Z are the same. Then the diagram $X \to Y \to \to R(Z)$ has the same property, so the two morphism to R(Z) are the same, and by adjunction the two maps to Z are the same. –  David Zureick-Brown Apr 2 '10 at 1:30