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 differential graded $\mathcal{O}_\mathcal {C}$-algebra. Let $\mathcal{B}$ be a differential graded $\mathcal{O}_\mathcal {D}$-algebra. Suppose we are given a map
of differential 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 differential graded $\mathcal{O}_\mathcal {C}$-algebras or equivalently $\varphi $ can be viewed as a map
of differential graded $\mathcal{O}_\mathcal {D}$-algebras. See Remark 24.12.2.
Let us define a functor
Given a differential graded $\mathcal{A}$-module $\mathcal{M}$ we define $f_*\mathcal{M}$ to be the graded $\mathcal{B}$-module constructed in Section 24.9 with differential given by the maps $f_*d : f_*\mathcal{M}^ n \to f_*\mathcal{M}^{n + 1}$. The construction is clearly functorial in $\mathcal{M}$ and we obtain our functor.
Let us define a functor
Given a differential graded $\mathcal{B}$-module $\mathcal{N}$ we define $f^*\mathcal{N}$ to be the graded $\mathcal{A}$-module constructed in Section 24.9. Recall that
Since $f^{-1}\mathcal{N}$ comes with the differentials $f^{-1}\text{d} : f^{-1}\mathcal{N}^ n \to f^{-1}\mathcal{N}^{n + 1}$ we can view this tensor product as an example of the tensor product discussed in Section 24.17 which provides us with a differential. The construction is clearly functorial in $\mathcal{N}$ and we obtain our functor $f^*$.
The functors $f_*$ and $f^*$ are readily enhanced to give functors of differential graded categories
which do the same thing on underlying objects and are defined by functoriality of the constructions on homogenous morphisms of degree $n$.
Lemma 24.18.1. In the situation above we have
\[ \mathop{\mathrm{Hom}}\nolimits _{\textit{Mod}^{dg}(\mathcal{B}, \text{d})}( \mathcal{N}, f_*\mathcal{M}) = \mathop{\mathrm{Hom}}\nolimits _{\textit{Mod}^{dg}(\mathcal{A}, \text{d})}( f^*\mathcal{N}, \mathcal{M}) \]
Proof. Omitted. Hints: This is true for the underlying graded categories by Lemma 24.9.1. A calculation shows that these isomorphisms are compatible with differentials. $\square$
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