Lemma 76.5.3. 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 \mathcal{O}_ Z-modules
Lemma 76.5.3. 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 \mathcal{O}_ Z-modules
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}')^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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