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.

Answer

Attribution
Source : Link , Question Author : windsheaf , Answer Author : Community