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

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). $\square$

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