The Stacks project

Lemma 85.6.1. In Situation 85.3.3. Let $\mathcal{O}$ be a sheaf of rings on $\mathcal{C}_{total}$. There is a canonical morphism of ringed topoi $g_ n : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ n), \mathcal{O}_ n) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{total}), \mathcal{O})$ agreeing with the morphism $g_ n$ of Lemma 85.3.5 on underlying topoi. The functor $g_ n^* : \textit{Mod}(\mathcal{O}) \to \textit{Mod}(\mathcal{O}_ n)$ has a left adjoint $g_{n!}$. For $\mathcal{G}$ in $\textit{Mod}(\mathcal{O}_ n)$-modules the restriction of $g_{n!}\mathcal{G}$ to $\mathcal{C}_ m$ is

\[ \bigoplus \nolimits _{\varphi : [n] \to [m]} f_\varphi ^*\mathcal{G} \]

where $f_\varphi : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ m), \mathcal{O}_ m) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ n), \mathcal{O}_ n)$ is the morphism of ringed topoi agreeing with the previously defined $f_\varphi $ on topoi and using the map $\mathcal{O}(\varphi ) : f_\varphi ^{-1}\mathcal{O}_ n \to \mathcal{O}_ m$ on sheaves of rings.

Proof. By Lemma 85.3.5 we have $g_ n^{-1}\mathcal{O} = \mathcal{O}_ n$ and hence we obtain our morphism of ringed topoi. By Modules on Sites, Lemma 18.41.1 we obtain the adjoint $g_{n!}$. To prove the formula for $g_{n!}$ we first define a sheaf of $\mathcal{O}$-modules $\mathcal{H}$ on $\mathcal{C}_{total}$ with degree $m$ component the $\mathcal{O}_ m$-module

\[ \mathcal{H}_ m = \bigoplus \nolimits _{\varphi : [n] \to [m]} f_\varphi ^*\mathcal{G} \]

Given a map $\psi : [m] \to [m']$ the map $\mathcal{H}(\psi ) : f_\psi ^{-1}\mathcal{H}_ m \to \mathcal{H}_{m'}$ is given on components by

\[ f_\psi ^{-1} f_\varphi ^*\mathcal{G} \to f_\psi ^* f_\varphi ^*\mathcal{G} \to f_{\psi \circ \varphi }^*\mathcal{G} \]

Since this map $f_\psi ^{-1}\mathcal{H}_ m \to \mathcal{H}_{m'}$ is $\mathcal{O}(\psi ) : f_\psi ^{-1}\mathcal{O}_ m \to \mathcal{O}_{m'}$-semi-linear, this indeed does define an $\mathcal{O}$-module (use Lemma 85.3.4). Then one proves directly that

\[ \mathop{\mathrm{Mor}}\nolimits _{\mathcal{O}_ n}(\mathcal{G}, \mathcal{F}_ n) = \mathop{\mathrm{Mor}}\nolimits _{\mathcal{O}}(\mathcal{H}, \mathcal{F}) \]

proceeding as in the proof of Lemma 85.3.5. Thus $\mathcal{H} = g_{n!}\mathcal{G}$ as desired. $\square$

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