The Stacks project

Lemma 7.28.1. Let $f : \mathcal{C} \to \mathcal{D}$ be a morphism of sites corresponding to the continuous functor $u : \mathcal{D} \to \mathcal{C}$. Let $V \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ and set $U = u(V)$. Then the functor $u' : \mathcal{D}/V \to \mathcal{C}/U$, $V'/V \mapsto u(V')/U$ determines a morphism of sites $f' : \mathcal{C}/U \to \mathcal{D}/V$. The morphism $f'$ fits into a commutative diagram of topoi

\[ \xymatrix{ \mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U) \ar[r]_{j_ U} \ar[d]_{f'} & \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \ar[d]^ f \\ \mathop{\mathit{Sh}}\nolimits (\mathcal{D}/V) \ar[r]^{j_ V} & \mathop{\mathit{Sh}}\nolimits (\mathcal{D}). } \]

Using the identifications $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U) = \mathop{\mathit{Sh}}\nolimits (\mathcal{C})/h_ U^\# $ and $\mathop{\mathit{Sh}}\nolimits (\mathcal{D}/V) = \mathop{\mathit{Sh}}\nolimits (\mathcal{D})/h_ V^\# $ of Lemma 7.25.4 the functor $(f')^{-1}$ is described by the rule

\[ (f')^{-1}(\mathcal{H} \xrightarrow {\varphi } h_ V^\# ) = (f^{-1}\mathcal{H} \xrightarrow {f^{-1}\varphi } h_ U^\# ). \]

Finally, we have $f'_*j_ U^{-1} = j_ V^{-1}f_*$.

Proof. It is clear that $u'$ is continuous, and hence we get functors $f'_* = (u')^ s = (u')^ p$ (see Sections 7.5 and 7.13) and an adjoint $(f')^{-1} = (u')_ s = ((u')_ p\ )^\# $. The assertion $f'_*j_ U^{-1} = j_ V^{-1}f_*$ follows as

\[ (j_ V^{-1}f_*\mathcal{F})(V'/V) = f_*\mathcal{F}(V') = \mathcal{F}(u(V')) = (j_ U^{-1}\mathcal{F})(u(V')/U) = (f'_*j_ U^{-1}\mathcal{F})(V'/V) \]

which holds even for presheaves. What isn't clear a priori is that $(f')^{-1}$ is exact, that the diagram commutes, and that the description of $(f')^{-1}$ holds.

Let $\mathcal{H}$ be a sheaf on $\mathcal{D}/V$. Let us compute $j_{U!}(f')^{-1}\mathcal{H}$. We have

\begin{eqnarray*} j_{U!}(f')^{-1}\mathcal{H} & = ((j_ U)_ p(u'_ p\mathcal{H})^\# )^\# \\ & = ((j_ U)_ pu'_ p\mathcal{H})^\# \\ & = (u_ p(j_ V)_ p\mathcal{H})^\# \\ & = f^{-1}j_{V!}\mathcal{H} \end{eqnarray*}

The first equality by unwinding the definitions. The second equality by Lemma 7.13.4. The third equality because $u \circ j_ V = j_ U \circ u'$. The fourth equality by Lemma 7.13.4 again. All of the equalities above are isomorphisms of functors, and hence we may interpret this as saying that the following diagram of categories and functors is commutative

\[ \xymatrix{ \mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U) \ar[r]_{j_{U!}} & \mathop{\mathit{Sh}}\nolimits (\mathcal{C})/h_ U^\# \ar[r] & \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \\ \mathop{\mathit{Sh}}\nolimits (\mathcal{D}/V) \ar[r]^{j_{V!}} \ar[u]^{(f')^{-1}} & \mathop{\mathit{Sh}}\nolimits (\mathcal{D})/h_ V^\# \ar[r] \ar[u]^{f^{-1}} & \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \ar[u]^{f^{-1}} } \]

The middle arrow makes sense as $f^{-1}h_ V^\# = (h_{u(V)})^\# = h_ U^\# $, see Lemma 7.13.5. In particular this proves the description of $(f')^{-1}$ given in the statement of the lemma. Since by Lemma 7.25.4 the left horizontal arrows are equivalences and since $f^{-1}$ is exact by assumption we conclude that $(f')^{-1} = u'_ s$ is exact. Namely, because it is a left adjoint it is already right exact (Categories, Lemma 4.24.5). Hence we only need to show that it transforms a final object into a final object and commutes with fibre products (Categories, Lemma 4.23.2). Both are clear for the induced functor $f^{-1} : \mathop{\mathit{Sh}}\nolimits (\mathcal{D})/h_ V^\# \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})/h_ U^\# $. This proves that $f'$ is a morphism of sites.

We still have to verify that $(f')^{-1}j_ V^{-1} = j_ U^{-1}f^{-1}$. To see this use the formula above and the description in Lemma 7.25.7. Namely, combined these give, for any sheaf $\mathcal{G}$ on $\mathcal{D}$, that

\[ j_{U!}(f')^{-1}j_ V^{-1}\mathcal{G} = f^{-1}j_{V!}j_ V^{-1}\mathcal{G} = f^{-1}(\mathcal{G} \times h_ V^\# ) = f^{-1}\mathcal{G} \times h_ U^\# = j_{U!}j_ U^{-1}f^{-1}\mathcal{G}. \]

Since the functor $j_{U!}$ induces an equivalence $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})/h_ U^\# $ we conclude. $\square$

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