The Stacks project

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

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

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

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). $\square$