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 $$f:X→Yf:X \to Y$$ be a proper morphism. In section 6.4. of Liu’s book he introduces the $$rr$$-dualizing sheaf $$ωf\omega_f$$ for $$ff$$ which satisfies
$$f∗HomOX(F,ωf)≅HomOY(Rrf∗F,OY)f_*\hom_{\ox}(\mathcal{F},\omega_f) \cong \hom_{\mathcal{O}_Y}(R^rf_* \mathcal{F},\mathcal{O}_Y )$$
for all quasi-coherent sheaves $$F\mathcal{F}$$ on $$XX$$.

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

Let $$XX$$ be a proper, one-dimensional scheme over the field $$kk$$ and let $$Y=P1kY = \mathbb{P}_k^1$$. Let $$ωf\omega_f$$ denote the 0-dualizing sheaf for $$ff$$. Let $$ZZ$$ denote an irreducible component of $$XX$$. Let $$j:Z→Xj: Z \to X$$ denote the corresponding closed immersion. Then $$f∘jf \circ j$$ is again finite and we denote its dualizing sheaf by $$ωf∘j\omega_{f \circ j}$$.

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

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.