Lemma 21.31.8. Let $f : X \to Y$ be a morphism of $\textit{LC}_{qc}$. Then there are commutative diagrams of topoi

\[ \vcenter { \xymatrix{ \mathop{\mathit{Sh}}\nolimits (\textit{LC}_{qc}/X) \ar[r]_{f_{qc}} \ar[d]_{\epsilon _ X} & \mathop{\mathit{Sh}}\nolimits (\textit{LC}_{qc}/Y) \ar[d]^{\epsilon _ Y} \\ \mathop{\mathit{Sh}}\nolimits (\textit{LC}_{Zar}/X) \ar[r]^{f_{Zar}} & \mathop{\mathit{Sh}}\nolimits (\textit{LC}_{Zar}/Y) } } \quad \text{and}\quad \vcenter { \xymatrix{ \mathop{\mathit{Sh}}\nolimits (\textit{LC}_{qc}/X) \ar[r]_{f_{qc}} \ar[d]_{a_ X} & \mathop{\mathit{Sh}}\nolimits (\textit{LC}_{qc}/Y) \ar[d]^{a_ Y} \\ \mathop{\mathit{Sh}}\nolimits (X) \ar[r]^ f & \mathop{\mathit{Sh}}\nolimits (Y) } } \]

with $a_ X = \pi _ X \circ \epsilon _ X$, $a_ Y = \pi _ X \circ \epsilon _ X$. If $f$ is proper, then $a_ Y^{-1} \circ f_* = f_{qc, *} \circ a_ X^{-1}$.

**Proof.**
The morphism of topoi $f_{qc}$ is the one from Sites, Lemma 7.25.8 which in our case comes from the continuous functor $Z/Y \mapsto Z \times _ Y X/X$, see Sites, Lemma 7.27.3. The diagram on the left commutes because the corresponding continuous functors compose correctly (see Sites, Lemma 7.14.4). The diagram on the right commutes because the one on the left does and because of part (5) of Lemma 21.31.7.

Proof of the final assertion. The reader may repeat the proof of part (7a) of Lemma 21.31.7; we will instead deduce this from it. As $\epsilon _{Y, *}$ is the identity functor on underlying presheaves, it reflects isomorphisms. The description in Lemma 21.31.6 shows that $\epsilon _{Y, *} \circ a_ Y^{-1} = \pi _ Y^{-1}$ and similarly for $X$. To show that the canonical map $a_ Y^{-1}f_*\mathcal{F} \to f_{qc, *}a_ X^{-1}\mathcal{F}$ is an isomorphism, it suffices to show that

\[ \pi _ Y^{-1}f_*\mathcal{F} = \epsilon _{Y, *}a_ Y^{-1}f_*\mathcal{F} \to \epsilon _{Y, *}f_{qc, *}a_ X^{-1}\mathcal{F} = f_{Zar, *}\epsilon _{X, *}a_ X^{-1}\mathcal{F} = f_{Zar, *}\pi _ X^{-1}\mathcal{F} \]

is an isomorphism. This is part (7a) of Lemma 21.31.7.
$\square$

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