Lemma 76.5.3. Let $S$ be a scheme. Let

$\xymatrix{ Z \ar[r]_ i \ar[d]_ f & X \ar[d]^ g \\ Z' \ar[r]^{i'} & X' }$

be a commutative diagram of algebraic spaces over $S$. Assume $i$, $i'$ immersions. There is a canonical map of $\mathcal{O}_ Z$-modules

$f^*\mathcal{C}_{Z'/X'} \longrightarrow \mathcal{C}_{Z/X}$

Proof. First find open subspaces $U' \subset X'$ and $U \subset X$ such that $g(U) \subset U'$ and such that $i(Z) \subset U$ and $i(Z') \subset U'$ are closed (proof existence omitted). Replacing $X$ by $U$ and $X'$ by $U'$ we may assume that $i$ and $i'$ are closed immersions. Let $\mathcal{I}' \subset \mathcal{O}_{X'}$ and $\mathcal{I} \subset \mathcal{O}_ X$ be the quasi-coherent sheaves of ideals associated to $i'$ and $i$, see Morphisms of Spaces, Lemma 67.13.1. Consider the composition

$g^{-1}\mathcal{I}' \to g^{-1}\mathcal{O}_{X'} \xrightarrow {g^\sharp } \mathcal{O}_ X \to \mathcal{O}_ X/\mathcal{I} = i_*\mathcal{O}_ Z$

Since $g(i(Z)) \subset Z'$ we conclude this composition is zero (see statement on factorizations in Morphisms of Spaces, Lemma 67.13.1). Thus we obtain a commutative diagram

$\xymatrix{ 0 \ar[r] & \mathcal{I} \ar[r] & \mathcal{O}_ X \ar[r] & i_*\mathcal{O}_ Z \ar[r] & 0 \\ 0 \ar[r] & g^{-1}\mathcal{I}' \ar[r] \ar[u] & g^{-1}\mathcal{O}_{X'} \ar[r] \ar[u] & g^{-1}i'_*\mathcal{O}_{Z'} \ar[r] \ar[u] & 0 }$

The lower row is exact since $g^{-1}$ is an exact functor. By exactness we also see that $(g^{-1}\mathcal{I}')^2 = g^{-1}((\mathcal{I}')^2)$. Hence the diagram induces a map $g^{-1}(\mathcal{I}'/(\mathcal{I}')^2) \to \mathcal{I}/\mathcal{I}^2$. Pulling back (using $i^{-1}$ for example) to $Z$ we obtain $i^{-1}g^{-1}(\mathcal{I}'/(\mathcal{I}')^2) \to \mathcal{C}_{Z/X}$. Since $i^{-1}g^{-1} = f^{-1}(i')^{-1}$ this gives a map $f^{-1}\mathcal{C}_{Z'/X'} \to \mathcal{C}_{Z/X}$, which induces the desired map. $\square$

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