Lemma 24.9.1. In the situation above we have
\mathop{\mathrm{Hom}}\nolimits _{\textit{Mod}^{gr}(\mathcal{B})}( \mathcal{N}, f_*\mathcal{M}) = \mathop{\mathrm{Hom}}\nolimits _{\textit{Mod}^{gr}(\mathcal{A})}( f^*\mathcal{N}, \mathcal{M})
24.9 Pull and push for sheaves of graded modules
We advise the reader to skip this section.
Let (f, f^\sharp ) : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}_\mathcal {C}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D}) be a morphism of ringed topoi. Let \mathcal{A} be a graded \mathcal{O}_\mathcal {C}-algebra. Let \mathcal{B} be a graded \mathcal{O}_\mathcal {D}-algebra. Suppose we are given a map
\varphi : f^{-1}\mathcal{B} \to \mathcal{A}
of graded f^{-1}\mathcal{O}_\mathcal {D}-algebras. By the adjunction of restriction and extension of scalars, this is the same thing as a map \varphi : f^*\mathcal{B} \to \mathcal{A} of graded \mathcal{O}_\mathcal {C}-algebras or equivalently \varphi can be viewed as a map
\varphi : \mathcal{B} \to f_*\mathcal{A}
of graded \mathcal{O}_\mathcal {D}-algebras. See Remark 24.3.2.
Let us define a functor
f_* : \textit{Mod}(\mathcal{A}) \longrightarrow \textit{Mod}(\mathcal{B})
Given a graded \mathcal{A}-module \mathcal{M} we define f_*\mathcal{M} to be the graded \mathcal{B}-module whose degree n term is f_*\mathcal{M}^ n. As multiplication we use
f_*\mathcal{M}^ n \times \mathcal{B}^ m \xrightarrow {(\text{id}, \varphi ^ m)} f_*\mathcal{M}^ n \times f_*\mathcal{A}^ m \xrightarrow {f_*\mu _{n, m}} f_*\mathcal{M}^{n + m}
where \mu _{n, m} : \mathcal{M}^ n \times \mathcal{A}^ m \to \mathcal{M}^{n + m} is the multiplication map for \mathcal{M} over \mathcal{A}. This uses that f_* commutes with products. The construction is clearly functorial in \mathcal{M} and we obtain our functor.
Let us define a functor
f^* : \textit{Mod}(\mathcal{B}) \longrightarrow \textit{Mod}(\mathcal{A})
We will define this functor as a composite of functors
\textit{Mod}(\mathcal{B}) \xrightarrow {f^{-1}} \textit{Mod}(f^{-1}\mathcal{B}) \xrightarrow { - \otimes _{f^{-1}\mathcal{B}} \mathcal{A}} \textit{Mod}(\mathcal{A})
First, given a graded \mathcal{B}-module \mathcal{N} we define f^{-1}\mathcal{N} to be the graded f^{-1}\mathcal{B}-module whose degree n term is f^{-1}\mathcal{N}^ n. As multiplication we use
f^{-1}\nu _{n, m} : f^{-1}\mathcal{N}^ n \times f^{-1}\mathcal{B}^ m \longrightarrow f^{-1}\mathcal{N}^{n + m}
where \nu _{n, m} : \mathcal{N}^ n \times \mathcal{B}^ m \to \mathcal{N}^{n + m} is the multiplication map for \mathcal{N} over \mathcal{B}. This uses that f^{-1} commutes with products. The construction is clearly functorial in \mathcal{N} and we obtain our functor f^{-1}. Having said this, we can use the tensor product discussion in Section 24.8 to define the functor
- \otimes _{f^{-1}\mathcal{B}} \mathcal{A} : \textit{Mod}(f^{-1}\mathcal{B}) \longrightarrow \textit{Mod}(\mathcal{A})
Finally, we set
f^*\mathcal{N} = f^{-1}\mathcal{N} \otimes _{f^{-1}\mathcal{B}, \varphi } \mathcal{A}
as already foretold above.
The functors f_* and f^* are readily enhanced to give functors of graded categories
f_* : \textit{Mod}^{gr}(\mathcal{A}) \longrightarrow \textit{Mod}^{gr}(\mathcal{B}) \quad \text{and}\quad f^* : \textit{Mod}^{gr}(\mathcal{B}) \longrightarrow \textit{Mod}^{gr}(\mathcal{A})
which do the same thing on underlying objects and are defined by functoriality of the constructions on homogeneous morphisms of degree n.
Proof. Omitted. Hints: First prove that f^{-1} and f_* are adjoint as functors between \textit{Mod}(\mathcal{B}) and \textit{Mod}(f^{-1}\mathcal{B}) using the adjunction between f^{-1} and f_* on sheaves of abelian groups. Next, use the adjunction between base change and restriction given in Section 24.8. \square
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