If one has a topological space $X$ and three presheaves resp. sheaves $F$ and $G$ and $H$ of abelian groups on it with morphisms of presheaves resp. sheaves $F\rightarrow H$, $G \rightarrow H$, then I wonder if one can consider a fiber product of $F$ and $G$ over $H$ in the category of presheaves and sheaves $F \times_HG$
Well, one would perhaps define it in the category of presheaves just as
$F\times _HG (U)$:= all (s,t) $\in F(U)\times G(U)$ going to the same element in $H(U)$.
And for sheaves then take the associated sheaf of this. Or perhaps it is already a sheaf if $F,G,H$ are sheaves?
Just give me some comment if this makes sense and if this concept is of relevance anywhere in Algebraic Geometry. I have never seen it indeed except for perhaps, if you want so, in the case of Schemes, where you consider a scheme as Zariski-Hom-sheaf.
Best Answer
Let $X$ be a topological space. The category of sheaves of sets on $X$, $\textrm{Sh}(X)$, is an example of a Grothendieck topos and is, in particular, complete and cocomplete. Therefore fibre products, or pullbacks as they are known in general category theory, exist in $\textrm{Sh}(X)$. Of course, one needs to verify that the forgetful functor $\textbf{Ab}(\textrm{Sh}(X)) \to \textrm{Sh}(X)$ preserves pullbacks, but this is straightforward.
Your proposed construction works and in fact does not require sheafification since limits and colimits of presheaves are constructed sectionwise, and the inclusion of categories $\textrm{Sh}(X) \hookrightarrow \textrm{Psh}(X)$ preserves all limits. (This is because it has a left adjoint, namely the associated sheaf functor.)
The fibre product of schemes is not an example of a fibre product of sheaves, unless you are thinking of a scheme as a sheaf in the Zariski topos.