The Stacks project

Lemma 7.27.5. Let $\mathcal{C}$ be a site. Let

\[ \xymatrix{ U' \ar[d] \ar[r] & U \ar[d] \\ V' \ar[r] & V } \]

be a commutative diagram of $\mathcal{C}$. The morphisms of Lemma 7.25.8 produce commutative diagrams

\[ \vcenter { \xymatrix{ \mathcal{C}/U' \ar[d]_{j_{U'/V'}} \ar[r]_{j_{U'/U}} & \mathcal{C}/U \ar[d]^{j_{U/V}} \\ \mathcal{C}/V' \ar[r]^{j_{V'/V}} & \mathcal{C}/V } } \quad \text{and}\quad \vcenter { \xymatrix{ \mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U') \ar[d]_{j_{U'/V'}} \ar[r]_{j_{U'/U}} & \mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U) \ar[d]^{j_{U/V}} \\ \mathop{\mathit{Sh}}\nolimits (\mathcal{C}/V') \ar[r]^{j_{V'/V}} & \mathop{\mathit{Sh}}\nolimits (\mathcal{C}/V) } } \]

of continuous and cocontinuous functors and of topoi. Moreover, if the initial diagram of $\mathcal{C}$ is cartesian, then we have $j_{V'/V}^{-1} \circ j_{U/V, *} = j_{U'/V', *} \circ j_{U'/U}^{-1}$.

Proof. The commutativity of the left square in the first statement of the lemma is immediate from the definitions. It implies the commutativity of the diagram of topoi by Lemma 7.21.2. Assume the diagram is cartesian. By the uniqueness of adjoint functors, to show $j_{V'/V}^{-1} \circ j_{U/V, *} = j_{U'/V', *} \circ j_{U'/U}^{-1}$ is equivalent to showing $j_{U/V}^{-1} \circ j_{V'/V!} = j_{U'/U!} \circ j_{U'/V'}^{-1}$. Via the identifications of Lemma 7.25.4 we may think of our diagram of topoi as

\[ \xymatrix{ \mathop{\mathit{Sh}}\nolimits (\mathcal{C})/h_{U'}^\# \ar[d] \ar[r] & \mathop{\mathit{Sh}}\nolimits (\mathcal{C})/h_ U^\# \ar[d] \\ \mathop{\mathit{Sh}}\nolimits (\mathcal{C})/h_{V'}^\# \ar[r] & \mathop{\mathit{Sh}}\nolimits (\mathcal{C})/h_ V^\# } \]

and we know how to interpret the functors $j^{-1}$ and $j_!$ by Lemma 7.25.9. Thus we have to show given $\mathcal{F} \to h_{V'}^\# $ that

\[ \mathcal{F} \times _{h_{V'}^\# } h_{U'}^\# = \mathcal{F} \times _{h_ V^\# } h_ U^\# \]

as sheaves with map to $h_ U^\# $. This is true because $h_{U'} = h_{V'} \times _{h_ V} h_ U$ and hence also

\[ h_{U'}^\# = h_{V'}^\# \times _{h_ V^\# } h_ U^\# \]

as sheafification is exact. $\square$

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