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Proof Let $n=\dim Y=\dim X$ and suppose $Y$ has rational singularities. Then $$ \omega_Y^{\bullet}\simeq R\mathscr Hom_Y(\mathscr O_Y, \omega_Y^{\bullet})\simeq R\mathscr Hom_Y(Rf_*\mathscr O_X, \omega_Y^{\bullet})\simeq Rf_*R\mathscr Hom_Y(\mathscr O_X, \omega_X^{\bullet})\simeq Rf_*\omega_X[n]\simeq f_*\omega_X[n]. $$ (The isomorphisms follow by the assumptions, Grothendieck duality, and the last one is the Grauert-Riemenschneider vanishing theorem). This implies that $\omega_Y=h^{-n}(\omega_Y^{\bullet})\simeq f_*\omega_X$, which is the second condition to prove and also that $h^i(\omega_Y^{\bullet})=0$ for $i\neq -n$ which is equivalent to $Y$ being Cohen-Macaulay.

Proof Let $n=\dim Y=\dim X$ and suppose $Y$ has rational singularities. Then $$ \omega_Y^{\bullet}\simeq R\mathscr Hom_Y(\mathscr O_Y, \omega_Y^{\bullet})\simeq R\mathscr Hom_Y(Rf_*\mathscr O_X, \omega_Y^{\bullet})\simeq Rf_*R\mathscr Hom_X(\mathscr O_X, \omega_X^{\bullet})\simeq Rf_*\omega_X[n]\simeq f_*\omega_X[n]. $$ (The isomorphisms follow by the assumptions, Grothendieck duality, and the last one is the Grauert-Riemenschneider vanishing theorem). This implies that $\omega_Y=h^{-n}(\omega_Y^{\bullet})\simeq f_*\omega_X$, which is the second condition to prove and also that $h^i(\omega_Y^{\bullet})=0$ for $i\neq -n$ which is equivalent to $Y$ being Cohen-Macaulay.

1 Elementary

2 High School

3 College

1 Easy

2 Less Easy

3 Even Less Easy

Proposition Let $f:X\to Y$ be a continuous map of topological spaces, $\mathscr F$ a sheaf of abelain groups on $X$ such that $R^jf_*\mathscr F=0$ for $j>0$. Then for all $i\geq 0$ there exists a natural isomorphism $$ H^i(Y, f_*\mathscr F)\simeq H^i(X,\mathscr F) $$

Furthermore, this proof shows more: Not only the cohomologies of these sheaves are isomorphic, but they come from the same complex! That's a much stronger statement. It is easy to give examples when the cohomologies of two complexes are isomoprhic, but the complexes are not. I suppose one may argue that the word "natural" in the statement means exactly this, but then I'd say that proving naturality with the Leray spectral sequence is certainly possible, but it definitely needs more care.

A special case of Grothedndieck duality says that if $f:X\to Y$ is a proper morphism between not too horrible schemes, let's say finite type over a field $k$ (let me not try to make a precise statement, this is in Residues and Duality that you mentioned) and $\mathscr F$ is a coherent sheaf on $X$, then $$ Rf_*R\mathscr Hom_X(\mathscr F, \omega_X^{\bullet})\simeq R\mathscr Hom_Y(Rf_*\mathscr F, \omega_Y^{\bullet}). $$ Here $\omega_{Z}^{\bullet}=\varepsilon^!k$ is "the" dulaizing complex where $\varepsilon: Z\to \mathrm{Spec}\ k$ is the structure map of $Z$.

Here is an application of Grothedieck duality where one can see how the derived category formalism makes life easier and arguments that seemed complicated are reduced to a one liner.

To answer your second question, I think you are right. In order to get the spectral sequences you do not need to go through the derived category formalism. However, if you are indeed "recovering" the spectral sequence, then you start with the derived category formalism. In other words, if you've never heard of derived catgories, why (or perhaps more importantly how) would you want to recover anything from a derived category statement? (Since written word lacks intonation, let me add that I'm not trying to be confrontational, but I feel that this question is somehow off target.)

Proposition Let $f:X\to Y$ be a continuous map of topological spaces, $\mathscr F$ a sheaf of abelian groups on $X$ such that $R^jf_*\mathscr F=0$ for $j>0$. Then for all $i\geq 0$ there exists a natural isomorphism $$ H^i(Y, f_*\mathscr F)\simeq H^i(X,\mathscr F) $$

Furthermore, this proof shows more: Not only the cohomologies of these sheaves are isomorphic, but they come from the same complex! That's a much stronger statement. It is easy to give examples when the cohomologies of two complexes are isomorphic, but the complexes are not. I suppose one may argue that the word "natural" in the statement means exactly this, but then I'd say that proving naturality with the Leray spectral sequence is certainly possible, but it definitely needs more care.

A special case of Grothendieck duality says that if $f:X\to Y$ is a proper morphism between not too horrible schemes, let's say finite type over a field $k$ (let me not try to make a precise statement, this is in Residues and Duality that you mentioned) and $\mathscr F$ is a coherent sheaf on $X$, then $$ Rf_*R\mathscr Hom_X(\mathscr F, \omega_X^{\bullet})\simeq R\mathscr Hom_Y(Rf_*\mathscr F, \omega_Y^{\bullet}). $$ Here $\omega_{Z}^{\bullet}=\varepsilon^!k$ is "the" dualizing complex where $\varepsilon: Z\to \mathrm{Spec}\ k$ is the structure map of $Z$.

Here is an application of Grothendieck duality where one can see how the derived category formalism makes life easier and arguments that seemed complicated are reduced to a one liner.

To answer your second question, I think you are right. In order to get the spectral sequences you do not need to go through the derived category formalism. However, if you are indeed "recovering" the spectral sequence, then you start with the derived category formalism. In other words, if you've never heard of derived categories, why (or perhaps more importantly how) would you want to recover anything from a derived category statement? (Since written word lacks intonation, let me add that I'm not trying to be confrontational, but I feel that this question is somehow off target.)