Equivalence of the definitions of a sheaf in SGA4 and in "Categories and Sheaves"

Question

I asked this question on Mathematics Stack Exchange, but got no answer.

I don't understand why the definition of a sheaf (Definition 17.3.1 (ii)) given in the book

[KS] Categories and Sheaves by Kashiwara and Schapira

is equivalent to the definition of a sheaf (Definition 2.1) given in

[V] Verdier, Exposé II, SGA4, http://www.normalesup.org/~forgogozo/SGA4/02/02.pdf

To simplify, let me consider only set-valued presheaves.

Here is, in the terminology of [KS], how I understand the two definitions. (Warning: my understanding might be incorrect!)

Let $\mathcal U$ be a universe, let $X$ be a small site and let $F$ be a $\mathcal U$-set-valued presheaf over $X$. Then:

• $F$ is a sheaf in the sense of [V] if $F(f)$ is an isomorphism for any $A$ in $(\mathcal C_X)^\wedge$, any $U$ in $\mathcal C_X$, and any local isomorphism $f:A\to U$ which is a monomorphism,

• $F$ is a sheaf in the sense of [KS] if $F(f)$ is an isomorphism for any $A$ in $(\mathcal C_X)^\wedge$, any $U$ in $\mathcal C_X$, and any local isomorphism $f:A\to U$.

The difference is that the local isomorphism $f$ is supposed to be a monomorphism in Verdier's definition.

A "KS-sheaf" is of course a "V-sheaf", but I'm unable to prove the converse.

Answer 1

$\DeclareMathOperator\im{im}$Actually I couldn't quite figure out how to do a Ken Brown sort of argument, but here's an argument that works:

Let $L$ denote the usual sheafification functor, a la Grothendieck and Verdier etc. Then I claim it's enough to show $L$ sends local epimorphisms to epimorphisms and local monomorphisms to monomorphisms. Indeed, if this is the case then $L$ takes local isomorphisms to epi-monomorphisms, and such things are isomorphisms in toposes.

Since $L$ preserves finite limits, we need only check that $L$ takes local epis to epis (since $A \to B$ is a local mono iff $A \to A\times_BA$ is a local epi). But that's not so bad: If $A \to B$ is a local epimorphism then the map $\im(A \to B) \to B$ is a local epimorphism and a monomorphism, and hence a monic, local isomorphism. But $L$ preserves images (since they're computed as colimits and L preserves those) and takes monic, local isomorphisms to isomorphisms. Thus $LA \to LB$ has the property that $\im(LA \to LB) \to LB$ is an isomorphism (NB: that image is computed in the category of sheaves), and so the map is epi in the category of sheaves, which is what we wanted.

Answer 2

I have found a very elementary proof, where the presheaves are allowed to take values in any category $\mathcal{A}$ with enough limits (however Dylan’s argument can be used to show this in the same generality by using Prop. 17.3.2 of KS). Let $F$ be a presheaf satisfying the Verdier definition. We shall first show that $F$ is separated in the sense of KS. For this let $f:A\rightarrow U$ be a local isomorphism. Factor $f=i\circ g$ with $g:A\rightarrow \operatorname{im}(f)$. Then $i$ is a monic local isomorphism and $g$ is epic. Consequently $F(i)$ is an isomorphism and as $F$ commutes with limits, $F(g)$ is a mono morphism. In particular $F(f)$ is Mono. This shows that $F$ is KS separated.

Now we shall show that $F(f)$ must even be an iso. For this, notice that as $f$ is a local monomorphism we also have that $g$ is a local monomorphism. Now as $g$ is a strict epimorphism (as we are in a category of sets), we get that $$A\times_{\operatorname{im}(f)}A\rightrightarrows A \rightarrow \operatorname{im}(f)$$ is a right exact sequence (i.e. $\operatorname{im}(f)$ is a cokernel of the two projections $p_1$ and $p_2$). As $F$ commutes with limits, we get the left exact sequence $$F(\operatorname{im}(f))\rightarrow F(A) \rightrightarrows F(A\times_{\operatorname{im}(f)}A).$$

Now as $g$ is a local monomorphism, we have that the diagonal map $\Delta:A\rightarrow A\times_{\operatorname{im}(f)}A $ is a local epimorphism. Hence, as $F$ is KS separated, proposition 17.3.3 implies that $F(\Delta)$ is monic. As $F(\Delta)\circ F(p_1)= F(\Delta)\circ F(p_2) $, we get $F(p_1)=F(p_2)$ and hence the last exact sequence forces $F(\operatorname{im}(f))\rightarrow F(A) $ to be an iso. But this morphism is $F(g)$. As $F(i)$ is an iso, so is $F(f)$ then.