Lemma 53.4.5. Let $X$ be a proper scheme of dimension $\leq 1$ over a field $k$. Let $i : Y \to X$ be a closed immersion. Let $\omega _ X^\bullet$, $\omega _ X$, $\omega _ Y^\bullet$, $\omega _ Y$ be as in Lemma 53.4.1. Then

1. $\omega _ Y^\bullet = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Y, \omega _ X^\bullet )$,

2. $\omega _ Y = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Y, \omega _ X)$ and $i_*\omega _ Y = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ Y, \omega _ X)$.

Proof. Denote $g : Y \to \mathop{\mathrm{Spec}}(k)$ and $f : X \to \mathop{\mathrm{Spec}}(k)$ the structure morphisms. Then $g = f \circ i$. Denote $a, b, c$ the right adjoint of Duality for Schemes, Lemma 48.3.1 for $f, g, i$. Then $b = c \circ a$ by uniqueness of right adjoints and because $Rg_* = Rf_* \circ Ri_*$. In the proof of Lemma 53.4.1 we set $\omega _ X^\bullet = a(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)})$ and $\omega _ Y^\bullet = b(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)})$. Hence $\omega _ Y^\bullet = c(\omega _ X^\bullet )$ which implies (1) by Duality for Schemes, Lemma 48.9.7. Since $\omega _ X = H^{-1}(\omega _ X^\bullet )$ and $\omega _ Y = H^{-1}(\omega _ Y^\bullet )$ we conclude that $\omega _ Y = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Y, \omega _ X)$. This implies $i_*\omega _ Y = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ Y, \omega _ X)$ by Duality for Schemes, Lemma 48.9.3. $\square$

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