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

Let f:XY be a proper morphism. In section 6.4. of Liu’s book he introduces the r-dualizing sheaf ωf for f which satisfies
fHomOX(F,ωf)HomOY(RrfF,OY)
for all quasi-coherent sheaves F on X.

In the special case of f being finite he proves that the 0-dualizing sheaf is given by f!OY=HomOY(fOX,OY) where this is considered an OX-module via multiplication into the argument.

Let X be a proper, one-dimensional scheme over the field k and let Y=P1k. Let ωf denote the 0-dualizing sheaf for f. Let Z denote an irreducible component of X. Let j:ZX denote the corresponding closed immersion. Then fj is again finite and we denote its dualizing sheaf by ωfj.

My question is: How is the restriction of ωf to Z (via the pullback j) related to the 0-dualizing sheaf ωfj for the morphism fj?

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

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

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