# How is the restriction of the dualizing sheaf to an irreducible component related to the dualizing sheaf of the component?

$$\DeclareMathOperator{\Spec}{Spec} \DeclareMathOperator{\hom}{\mathcal{Hom}} \DeclareMathOperator{\ox}{\mathcal{O}_X}$$Let $$f:X \to Y$$ be a proper morphism. In section 6.4. of Liu's book he introduces the $$r$$-dualizing sheaf $$\omega_f$$ for $$f$$ which satisfies $$f_*\hom_{\ox}(\mathcal{F},\omega_f) \cong \hom_{\mathcal{O}_Y}(R^rf_* \mathcal{F},\mathcal{O}_Y )$$ for all quasi-coherent sheaves $$\mathcal{F}$$ on $$X$$.

In the special case of $$f$$ being finite he proves that the 0-dualizing sheaf is given by $$f^!\mathcal{O}_Y = \hom_{\mathcal{O}_Y}(f_*\mathcal{O}_X,\mathcal{O}_Y)$$ where this is considered an $$\mathcal{O}_X$$-module via multiplication into the argument.

Let $$X$$ be a proper, one-dimensional scheme over the field $$k$$ and let $$Y = \mathbb{P}_k^1$$. Let $$\omega_f$$ denote the 0-dualizing sheaf for $$f$$. Let $$Z$$ denote an irreducible component of $$X$$. Let $$j: Z \to X$$ denote the corresponding closed immersion. Then $$f \circ j$$ is again finite and we denote its dualizing sheaf by $$\omega_{f \circ j}$$.

My question is: How is the restriction of $$\omega_f$$ to $$Z$$ (via the pullback $$j^*$$) related to the 0-dualizing sheaf $$\omega_{f \circ j}$$ for the morphism $$f \circ j$$?

Are they isomorphic? In general, I don't think so. But maybe in specific cases they are. Do there exist canonical maps between them?

I am grateful for any kind of help.

• You already wrote the answer in your post: $j_*\omega_{f\circ j}$ equals $\textit{Hom}_{\mathcal{O}_X}(j_*\mathcal{O}_Z,\omega_f)$ considered as a $j_*\mathcal{O}_Z$-module. The homomorphism $\mathcal{O}_X\to j_*\mathcal{O}_Z$ induces a homomorphism $j_*\omega_{f\circ j} \to \omega_f$. Typically these are not isomorphic. For $Z$ equal to a union of irreducible components of a strict normal crossings scheme, $j^*\omega_f$ is the "twist up" of $\omega_{f\circ j}$ by the Cartier divisor $D = Z\cap \overline{X\setminus Z}$. – Jason Starr Nov 27 '18 at 12:22
• @JasonStarr Do you know any geometric criteria on which the relation depends? For instance, is the canonical morphism an isomorphism if the irreducible components only intersect transversally? I would really like to get a 'feeling' for the situation. – windsheaf Nov 27 '18 at 15:48
• "For instance, is the canonical morphism an isomorphism if ..." No, it is usually not an isomorphism, even if there are just two irreducible components that intersect transversally in a divisor. You still need to "twist up" by the divisor before you get an isomorphism of $j^*\omega_f$ and $\omega_{f\circ j}$, i.e., $j^*\omega_f$ is isomorphic to $\omega_{f\circ j}(\underline{D})$. – Jason Starr Nov 27 '18 at 16:44