Sheaf on the big site $\operatorname{Sch}/S$ induced by sheaf on $S$

Question

On the Stacks Project it is claimed that, given a scheme $S$, any sheaf of modules $\mathcal F$ on $S$ induces a functor $$\operatorname{Sch}/S^{\mathrm{op}}\to \operatorname{Ab}$$ by sending $h\colon T\to S$ to $\Gamma(T,h^\ast F)$. Now I am having trouble believing that this is really a functor: Is it not true that for two morphisms $f\colon X\to Y$ and $g\colon Y\to Z$ of schemes one has $(g\circ f)^\ast \simeq f^\ast\circ g^\ast$ and not strict equality? After all, the assignment $(T\to S)\mapsto \operatorname{Mod}(T)$ is a pseudofunctor $$\operatorname{Sch}/S^{\mathrm{op}}\to \operatorname{Cat}$$ and not a functor.

Answer 1

As you said, the assignment $(T\to S)\mapsto\operatorname{Mod}(T)$ is a pseudo-functor, but this implies that $(h:T\to S)\mapsto \Gamma(T,h^*F)$ is a functor. Indeed, if $u$ is a map from $h:T\to S$ to $h':T'\to S$ (in other words that there is a commutative triangle) then there is a well-defined morphism : $$u^*:\Gamma(T',h'^*F)\to\Gamma(T',u_*u^*h'^*F)=\Gamma(T,u^*h'^*F)\simeq\Gamma(T,(h'u)^*F)=\Gamma(T,h^*F) $$ The point is that morphism represented by $\simeq$ (which as you said is not an identity) is canonical and satisfy a bunch of properties which allow to prove that

• $\mathrm{id}_{h:T\to S}^*=\mathrm{id}_{\Gamma(T,h^*F)}$
• $(uv)^*=v^*u^*$ as morphisms between two groups.

To prove the last statement, you will need to use the pentagonal diagram relating $(v^*u^*)h'^*, v^*(u^*h'^*), (uv)^*h'^*, v^*(h'u)^*, (h'uv)^*$. This is quite long, and I won't do it here.