Lemma 29.31.3. Let
\xymatrix{ Z \ar[r]_ i \ar[d]_ f & X \ar[d]^ g \\ Z' \ar[r]^{i'} & X' }
be a commutative diagram in the category of schemes. Assume i, i' immersions. There is a canonical map of \mathcal{O}_ Z-modules
f^*\mathcal{C}_{Z'/X'} \longrightarrow \mathcal{C}_{Z/X}
characterized by the following property: For every pair of affine opens (\mathop{\mathrm{Spec}}(R) = U \subset X, \mathop{\mathrm{Spec}}(R') = U' \subset X') with f(U) \subset U' such that Z \cap U = \mathop{\mathrm{Spec}}(R/I) and Z' \cap U' = \mathop{\mathrm{Spec}}(R'/I') the induced map
\Gamma (Z' \cap U', \mathcal{C}_{Z'/X'}) = I'/I'^2 \longrightarrow I/I^2 = \Gamma (Z \cap U, \mathcal{C}_{Z/X})
is the one induced by the ring map f^\sharp : R' \to R which has the property f^\sharp (I') \subset I.
Proof.
Let \partial Z' = \overline{Z'} \setminus Z' and \partial Z = \overline{Z} \setminus Z. These are closed subsets of X' and of X. Replacing X' by X' \setminus \partial Z' and X by X \setminus \big (g^{-1}(\partial Z') \cup \partial Z\big ) we see that we may assume that i and i' are closed immersions.
The fact that g \circ i factors through i' implies that g^*\mathcal{I}' maps into \mathcal{I} under the canonical map g^*\mathcal{I}' \to \mathcal{O}_ X, see Schemes, Lemmas 26.4.6 and 26.4.7. Hence we get an induced map of quasi-coherent sheaves g^*(\mathcal{I}'/(\mathcal{I}')^2) \to \mathcal{I}/\mathcal{I}^2. Pulling back by i gives i^*g^*(\mathcal{I}'/(\mathcal{I}')^2) \to i^*(\mathcal{I}/\mathcal{I}^2). Note that i^*(\mathcal{I}/\mathcal{I}^2) = \mathcal{C}_{Z/X}. On the other hand, i^*g^*(\mathcal{I}'/(\mathcal{I}')^2) = f^*(i')^*(\mathcal{I}'/(\mathcal{I}')^2) = f^*\mathcal{C}_{Z'/X'}. This gives the desired map.
Checking that the map is locally described as the given map I'/(I')^2 \to I/I^2 is a matter of unwinding the definitions and is omitted. Another observation is that given any x \in i(Z) there do exist affine open neighbourhoods U, U' with f(U) \subset U' and Z \cap U as well as U' \cap Z' closed such that x \in U. Proof omitted. Hence the requirement of the lemma indeed characterizes the map (and could have been used to define it).
\square
Comments (1)
Comment #9825 by George Cooper on