Lemma 18.40.11. If $f : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}') \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ is a morphism of topoi. If $\mathcal{O}$ is a sheaf of rings on $\mathcal{C}$, then
\[ f^{-1}(\mathcal{O}^*) = (f^{-1}\mathcal{O})^*. \]
In particular, if $\mathcal{O}$ turns $\mathcal{C}$ into a locally ringed site, then setting $f^\sharp = \text{id}$ the morphism of ringed topoi
\[ (f, f^\sharp ) : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), f^{-1}\mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}, \mathcal{O}) \]
is a morphism of locally ringed topoi.
Proof.
Note that the diagram
\[ \xymatrix{ \mathcal{O}^* \ar[rr] \ar[d]_{u \mapsto (u, u^{-1})} & & {*} \ar[d]^{1} \\ \mathcal{O} \times \mathcal{O} \ar[rr]^-{(a, b) \mapsto ab} & & \mathcal{O} } \]
is cartesian. Since $f^{-1}$ is exact we conclude that
\[ \xymatrix{ f^{-1}(\mathcal{O}^*) \ar[d]_{u \mapsto (u, u^{-1})} \ar[rr] & & {*} \ar[d]^{1} \\ f^{-1}\mathcal{O} \times f^{-1}\mathcal{O} \ar[rr]^-{(a, b) \mapsto ab} & & f^{-1}\mathcal{O} } \]
is cartesian which implies the first assertion. For the second, note that $(\mathcal{C}', f^{-1}\mathcal{O})$ is a locally ringed site by Lemma 18.40.5 so that the assertion makes sense. Now the first part implies that the morphism is a morphism of locally ringed topoi.
$\square$
Comments (1)
Comment #8817 by Jakob Werner on