# Does a sheaf of abelian groups on a scheme $X$ induce a sheaf of abelian groups on the étale site $X_{ét}$?

Fixed a scheme $$X$$, étale cohomology $$H^\bullet(X,F)$$ is defined for all sheaves of abelian groups $$F$$ over the étale site $$X_{ét}$$.

Now, just to understand, I tried to see if this covers sheaves of abelian groups over $$X$$ itself as well (i.e., sheaves over the Zariski site). So, does a sheaf of abelian groups on a scheme $$X$$ induce a sheaf of abelian groups on the étale site $$X_{ét}$$, for which the étale cohomology groups are defined?

My guess is that, given a sheaf of abelian groups $$F$$ over $$X$$, the functor $$\tilde F:X_{ét}\to \mathbf{AbGrp}$$ defined by $$\tilde F(g:Y\to X)=(g^*F)(X)$$ is a sheaf of abelian groups over $$X_{ét}$$.

Is this true? Is this the natural way how sheaves over $$X$$ become handleable by étale cohomology, if there is any?

Thank you in advance.

Let $$F$$ be a sheaf of Abelian Groups on the Zariski site. Let us extend the definition as described in the question of the Zariski sheaf $$F$$ to a functor defined on the etale site and call it $$F_{et}$$. We first verify that this is indeed a sheaf.

First certain preliminaries.

Let $$g : U \rightarrow X$$ be an etale open set in $$X$$, then $$F_{et}(g:U \rightarrow X) = F(g(U))$$. This is easy to see by the definition of the pullback of a sheaf and the fact that $$g(U) \subset X$$ is an open subset in the zariski topology, since $$g$$ is etale in particular flat. Next, let us see what happens on the intersection of two such open sets.

Let $$g : U \rightarrow X$$ and $$h : V \rightarrow X$$ be two open subsets. Let $$g\times_X h : U \times_X V \rightarrow X$$ be the fiber product. Then $$F_{et}(g \times_X h : U \times_X V \rightarrow X) = F((g \times_X h)(U \times_X V \rightarrow X)) = F(g(U) \cap h(V))$$. The first equality follows as above while the second equality follows from the fact that $$(g \times_X h)(U \times_X V \rightarrow X) = g(U) \cap h(V)$$. This is explained below :

We have the following setup $$U \rightarrow g(U) \hookrightarrow X$$, where the first arrow is surjective and the second is an open immersion(topologically and if we give (and we do) $$g(U)$$ the induced subscheme structure, then scheme theoretically). A similar factorization exists for the arrow $$h$$. Using the universal property of cartesian diagrams we get that $$g \times_X h$$ factors through $$g(U) \times_X h(U) = g(U) \cap h(V)$$(This is clear since $$g(U)$$ and $$h(V)$$ are subsets of $$X$$). Let us analyze the map $$U \times_X V \rightarrow g(U) \times_X h(U)$$.

Claim : The above map $$U \times_X V \rightarrow g(U) \times_X h(U)$$ is surjective.

Proof of Claim : Note that this map factors as follows $$U \times_X V \rightarrow g(U)\times_X V \rightarrow g(U) \times_X h(U)$$. Now, note that $$U \rightarrow g(U)$$ and $$V \rightarrow h(V)$$ is surjective. We know that base change of a surjective morphism is surjective. Hence both the arrows are surjective maps and hence the composition is surjective. Thus proving the claim.

Now, we prove that $$F_{et}$$ is infact a sheaf. Let $$\lbrace g_i : U_i \rightarrow U \rbrace$$ be an etale cover of an etale open set $$g : U \rightarrow X$$ i.e. each of the $$g_i$$ is an etale map and $$\cup_i g_i(U_i) = g(U)$$. We consider the standard sequence of Abelian groups

$$F_{et}(g:U \rightarrow X) \rightarrow \prod^i F_{et}(g_i) \rightarrow \prod F_{et}(g_i \times_X g_j)$$

From above paragraphs, the exact sequence is same as the exact sequence

$$F(g(U)) \rightarrow \prod F(g_i(U_i)) \rightarrow F(g_i(U_i) \cap g_j(U_j))$$

This is an exact sequence follows from the fact that $$F$$ is a sheaf on the zariski site.

Thus we have that $$F_{et}$$ is infact a sheaf. Note that the etale cohomology of this sheaf will be the same as usual cohomology as can be seen by considering cech complexes.

This is infact functorial for the morphism for sheaves.