A question about how we get from PreSheaves of X to those of Y - why is $f_{*}F(V) = F(f^{t}(V)) \in PSh(k_Y)$

23 Views Asked by At

The problem that I am struggling with is the meanings and relationships of $f_{*}, f^{t}, f^{-1}, \text{and} f^{\dagger}$. These definitions come from Shapirra's notes on Algebra and Topology which can be found here: http://www.math.jussieu.fr/~schapira/lectnotes/AlTo.pdf . Most of which are defined on P103. First, a few definitions from his notes.

Let $f: X \rightarrow Y$ be a continuous map. We denote by $f^{t}$ the inverse image of a set by $f$. Hence, we set for $V \subset Y: f^t(V) := f^{-1}(V)$ and $f^t: Op_Y \rightarrow Op_X$ is a functor.

Let $f: X \rightarrow Y$ be a morphism of sites and let $F \in PSh(k_X)$. One defines $f_{*}F \in PSh(k_Y)$, the direct image of $F$ by $f$, by setting $f_{*}F(V) = F(f^{t}(V))$.

I can see that $f^t(V)$ gives us a set in $Op_X$ which is the set of open sets of $X$. Now since we have $f^t$ operating on $V \subset Y$ and end up with an open set of $X$, how it is that the action of $F$ on this open set of $X$ gives us an element of $Y$ again i.e. why is $f_{*}F(V) \in PSh(k_Y)$?

The author goes on to say: Proposition. Let $F \in Mod(k_X)$. Then $f_{*} \in Mod(k_Y)$. In other words, the direct image of a sheaf is a sheaf.

Any thoughts?

Thank you,

Brian