If $X$ is a topological space, $\mathcal{F},\mathcal{G}$ are sheaves on $X$ with values in an abelian category $\mathcal{C}$, and $f:\mathcal{F}\to\mathcal{G}$ is a morphism of sheaves, then we can define a sheaf $\operatorname{coker}f$ as the sheafification of the presheaf defined by $U\mapsto\operatorname{coker}f_U$, where $U\subseteq X$ is open and $f_U:\mathcal{F}(U)\to\mathcal{G}(U)$ is the map on sections over $U$. $f$ is then called surjective if $\operatorname{coker}f$ is the zero sheaf. It is known (see here e.g.) that $f$ being surjective does not mean that all the $f_U$ have to be surjective, although it is true for the maps on the stalks.
I recently stumbled upon the Stacks Project and just out of curiosity I had a look at one of the chapters, namely the chapter on Sites and Sheaves. A 'categorical' presheaf is defined as a functor $\mathcal{F}:\mathcal{C}^\circ\to\mathbf{Set}$ for some category $\mathcal{C}$, and morphisms between presheaves as natural transformations between such functors. They say that a morphism $f$ of presheaves is surjective if the corresponding maps on the sections are surjective (Def. 3.1).
My question is: how do these two definitions go together? I know they don't really overlap, since 'categorical' presheaves take values in $\mathbf{Set}$, but I feel like there must be some connection at least. I posted a similar question some days ago, but it was full of errors, hence I deleted it again. Some things to make my points of confusion a little clearer, hopefully:
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In the first paragraph, does it even make sense to define surjectivity of presheaf morphisms via the cokernel presheaf, or is it only good for actual sheaves (since I never saw a definition for presheaves).
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If we take $\mathcal{C}$ in the second paragraph to be the category of open subsets of some topological space $X$, $\mathbf{Top}(X)$, we could talk about presheaves on $X$ with values in $\mathbf{Set}$ (which is not abelian of course). An 'ordinary' (pre)sheaf of, say, $k$-algebras on $X$ could be considered as such a presheaf. Now if $f$ is a morphism of sheaves on a topological space $X$ which is surjective in the first sense needn't be surjective in the sense of 'categorical' presheaves on $\mathbf{Top}(X)$, or do I have an error in there anywhere?
I'd be glad if maybe someone could elaborate on the connection between the notions of presheaf and surjective presheaf morphism in those two different senses. Thank you very much! (Note: I will be away for some days, so I most likely won't be able to react to comments / answers immediately, but I should be back on monday).
Best Answer
That categorical definition is for pre-sheaves, the topological definition is for sheaves.
In topological pre-sheaves, a map is surjective if it is epimorphic for each open set $U$ in $X$.
In topological sheaves, however, we instead have to "sheaf-ify" the definition, and we say that the map is "surjective" if the sheaf-ification of the cokernel map is zero.
Basically, in both cases, you have two categories, $\mathcal{Sh}$ and $\mathcal{PSh}$, and in $\mathcal{PSh}$, the "surjective" maps are the ones that are epimorphisms on each $U$, but in the $\mathcal{Sh}$ catageory, you have a more complicated definition of "surjective" (or "epimorphism.")
Consider, instead, two categories, $\mathcal{Ab}$ the category of abelian groups, and $\mathcal{AbTF}$, the full subcategory of "torsion-free" abelian groups - that is, the abelian groups, $A$, where for any $n\in\mathbb Z$ and $a\in A$, $na=0$ iff $n=0$ or $a=0$.
There is the natural inclusion functor $\mathcal{AbTF}\to\mathcal{Ab}$ and a natural adjoint sending $A\to A/N(A)$ where $N(A)$ is the subgroup of nilpotent elements of $A$.
But in $\mathcal{AbTF}$, the "epimorphisms" are not the ones with cokernel (in $\mathcal{Ab}$) $0$, they are the ones with cokerkels which are nilpotent. So, for example, in $\mathcal{Ab}$, the morphism $\mathbb Z\to\mathbb Z$ sending $x\to 2x$ is not an epimorphism, that same map, when considered as a map in $\mathcal{AbTF}$, is an epimorphism.
So consider the "sheafification" functor $\mathcal{PSh}\to \mathcal{Sh}$ to be much like the functor $\mathcal{Ab}\to\mathcal{AbTF}$.
(I believe, but don't quote me, that $f:A\to B$ in $\mathcal{AbTF}$ is an epimorphism if and only if $f\otimes \mathbb Q:A\otimes \mathbb Q\to B\otimes\mathbb Q$ is an epimorphism in $\mathcal{Ab}$.)