# geometry ag.algebraica: How is the restriction of the dualized sheaf to an irreducible component related to the dualizing sheaf of the component?

$$DeclareMathOperator { Spec} {Spec} DeclareMathOperator { hom} { mathcal {Hom}} DeclareMathOperator { ox} { mathcal {O} _X}$$Leave $$f: X to Y$$ Be your own map In section 6.4. from Liu's book presents the $$r$$-dualizing the sheaf $$omega_f$$ for $$f$$ what 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 $$mathcal {F}$$ in $$X$$.

In the special case of $$f$$ being finite, it shows that the sheaf of dual dualization is given by $$f ^! mathcal {O} _Y = hom_ { mathcal {O} _Y} (f _ * mathcal {O} _X, mathcal {O} _Y)$$ where this is considered a $$mathcal {O} _X$$-module by multiplication in the argument.

Leave $$X$$ To be a suitable one-dimensional scheme on the field. $$k$$ and let $$Y = mathbb {P} _k ^ 1$$. Leave $$omega_f$$ denote the dualization sheaf 0 for $$f$$. Leave $$Z$$ denotes an irreducible component of $$X$$. Leave $$j: Z to X$$ We denote the corresponding closed immersion. So $$f circ j$$ Again it is finite and we denote its sheaf dualized by $$omega_ {f circ j}$$.

My question is: How is the restriction of $$omega_f$$ to $$Z$$ (through the withdrawal $$j ^ *$$) related to the dualization sheaf 0 $$omega_ {f circ j}$$ by the morphism $$f circ j$$?

Are they isomorphic? In general, I do not think so. But maybe in specific cases they are. Are there canonical maps between them?

I'm grateful for any kind of help.