Lemma 53.23.4. Let $c : X \to Y$ be the contraction of a rational bridge (Example 53.23.1). Then $c^*\omega _ Y \cong \omega _ X$.

Proof. You can prove this by direct computation, but we prefer to use the characterization of $\omega _ X$ as the coherent $\mathcal{O}_ X$-module which represents the functor $\textit{Coh}(\mathcal{O}_ X) \to \textit{Sets}$, $\mathcal{F} \mapsto \mathop{\mathrm{Hom}}\nolimits _ k(H^1(X, \mathcal{F}), k) = H^1(X, \mathcal{F})^\vee$, see Lemma 53.4.2 or Duality for Schemes, Lemma 48.22.5.

To be precise, denote $\mathcal{C}_ Y$ the category whose objects are invertible $\mathcal{O}_ Y$-modules and whose maps are $\mathcal{O}_ Y$-module homomorphisms. Denote $\mathcal{C}_ X$ the category whose objects are invertible $\mathcal{O}_ X$-modules $\mathcal{L}$ with $\mathcal{L}|_ C \cong \mathcal{O}_ C$ and whose maps are $\mathcal{O}_ Y$-module homomorphisms. We claim that the functor

$c^* : \mathcal{C}_ Y \to \mathcal{C}_ X$

is an equivalence of categories. Namely, by More on Morphisms, Lemma 37.72.8 it is essentially surjective. Then the projection formula (Cohomology, Lemma 20.54.2) shows $c_*c^*\mathcal{N} = \mathcal{N}$ and hence $c^*$ is an equivalence with quasi-inverse given by $c_*$.

We claim $\omega _ X$ is an object of $\mathcal{C}_ X$. Namely, we have a short exact sequence

$0 \to \omega _ C \to \omega _ X|_ C \to \mathcal{O}_{C \cap X'} \to 0$

See Lemma 53.4.6. Taking degrees we find $\deg (\omega _ X|_ C) = 0$ (small detail omitted). Thus $\omega _ X|_ C$ is trivial by Lemma 53.10.1 and $\omega _ X$ is an object of $\mathcal{C}_ X$.

Since $R^1c_*\mathcal{O}_ X = 0$ the projection formula shows that $R^1c_*c^*\mathcal{N} = 0$ for $\mathcal{N} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_ Y)$. Therefore the Leray spectral sequence (Cohomology, Lemma 20.13.6) the diagram

$\xymatrix{ \mathcal{C}_ Y \ar[rr]_{c^*} \ar[dr]_{H^1(Y, -)^\vee } & & \mathcal{C}_ X \ar[ld]^{H^1(X, -)^\vee } \\ & \textit{Sets} }$

of categories and functors is commutative. Since $\omega _ Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_ Y)$ represents the south-east arrow and $\omega _ X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_ X)$ represents the south-east arrow we conclude by the Yoneda lemma (Categories, Lemma 4.3.5). $\square$

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