Presheaf kernel is a presheaf/sheaf

Question

$\mathscr F$ and $\mathscr G$ are assumed to be presheaves of abelian groups, i.e., contravariant functors from the category of open subsets of $X$ to the category of abelian groups. So $\mathscr F(U)$ and $\mathscr G(U)$ are abelian groups (in particular, sets). And we want to show that $\ker_{pre}(\phi)$ is a presheaf of abelian groups; in particular, $\ker_{pre}(\phi)(U)$ should be an abelian group.

So can I just define the dotted line from the diagram by sending an element $e$ to the same thing in $\mathscr F(V)$, and then sending it to $res_{V,U}(e)$? $res_{V,U}(e)$ indeed lies where we want it to lie because the rightmost square commutes.

I don't see anything wrong with this, but there's a similar qustion where (almost) the same problem was solved using general facts about abelian categories.

And assuming my definition is correct, if we assume that $\mathscr F,\mathscr G$ are sheaves, is it obvious that the kernel presheaf is a sheaf? The dotted restriction map is the same as the restriction map on $\mathscr F$ (middle vertical arrow), it's just that its domain and range is restricted. And we know that e.g. whenever $U$ is covered by the $U_i$ and $U_i\subset U$, then $res_{U,U_i}^{\mathscr F}(f_1)=res_{U,U_i}^{\mathscr F}(f_2)$ implies $f_1=f_2$, and hence the same holds for the restriction maps on $\ker_{pre}\phi$.

Answer 1

What you say here and what is written in the answer you refer to is exactly the same thing. The expression $$f(U)\circ \text{res}_{V,U}\circ \phi = \text{res}_{V,U}\circ \underbrace{f(V)\circ \phi}_{0} = 0$$ is your "the rightmost square commutes" and the conclusion that follows, that $\text{res}_{V,U}\circ \phi$ factors through $\ker_{\text{pre}}f(U),$ is exactly the same thing as saying, as you do, that $\text{res}_{V,U}\circ \phi(e)$ lies in the kernel.

Now, if $\mathscr F,\mathscr G$ are sheaves, the kernel is a sheaf indeed, and it can be proved the way you do. Indeed, if $\mathscr{F}$ is separated (i.e. there is at most one amalgamation in $\mathscr{F}$) and $K \hookrightarrow \mathscr{F}$, then $K$ is also separated. But beware that an $\mathscr{F}$-amalgamation of elements of $K$ might not be in $K$ anymore if $K$ is just some random subpresheaf (can you think of an example?). What you'd have to show is that in your case the amalgamation will be mapped to $0$ again.