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.