Lemma 76.6.3. Let $S$ be a scheme. Let
be a commutative diagram of algebraic spaces over $S$. Assume $i$, $i'$ immersions. There is a canonical map of graded $\mathcal{O}_ Z$-algebras
Lemma 76.6.3. Let $S$ be a scheme. Let
be a commutative diagram of algebraic spaces over $S$. Assume $i$, $i'$ immersions. There is a canonical map of graded $\mathcal{O}_ Z$-algebras
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
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
The lower row is exact since $g^{-1}$ is an exact functor. By exactness we also see that $(g^{-1}\mathcal{I}')^ n = g^{-1}((\mathcal{I}')^ n)$ for all $n \geq 1$. Hence the diagram induces a map $g^{-1}((\mathcal{I}')^ n/(\mathcal{I}')^{n + 1}) \to \mathcal{I}^ n/\mathcal{I}^{n + 1}$. Pulling back (using $i^{-1}$ for example) to $Z$ we obtain $i^{-1}g^{-1}((\mathcal{I}')^ n/(\mathcal{I}')^{n + 1}) \to \mathcal{C}_{Z/X, n}$. Since $i^{-1}g^{-1} = f^{-1}(i')^{-1}$ this gives maps $f^{-1}\mathcal{C}_{Z'/X', n} \to \mathcal{C}_{Z/X, n}$, which induce the desired map. $\square$
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