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## 1 Easy

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 $$H^i(Y, f_*\mathscr F)\simeq H^i(X,\mathscr F)$$

Proof Apply the composition rule for the derived functors of $G=\Gamma(Y, \_ )$ and $F=f_*({\_})$. By definition, $G\circ F = \Gamma(X, \_ )$. Then $$R\Gamma(Y, f_*\mathscr F) \simeq R\Gamma(Y, Rf_*\mathscr F) \simeq R\Gamma (X, \mathscr F).$$ Taking cohomology shows the result. $\square$

(edit to please anon, see the comments below) This is usually exhibited as an example of how to use the Leray spectral sequence. Doing it that way is not much harder than the above, but perhaps a bit less "automatic".

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.

This last point is actually an important one regarding the derived category language. You get a higher level notion. The fact that you can work with the complex whose cohomologies give the derived functors of your original functor is very very useful.

## 2 Less Easy

In case you were not convinced by the above example, here is one that should do the trick:

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 $Z$.

Now try to imagine how one could state this using spectral sequences. Both sides actually correspond to spectral sequences, so the statement would be something like "there is a natural map between this an this spectral sequences, such that they converge to the same thing".

I would argue that already the statement of this theorem would be tiring in the language of spectral sequences, but using it would be pure pain.

## 3 Even Less Easy

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.

Theorem (a.k.a. Kempf's Criterion) Let $Y$ be a normal variety over $\mathbb C$ with a resolution of singularities $f:X\to Y$. Then $Y$ has rational singularities (i.e., $R^if_*\mathscr O_X=0$ for $i>0$) if and only if

1. $Y$ is Cohen-Macaulay
2. $f_*\omega_X\simeq \omega_Y$

Proof Let $n=\dim Y=\dim X$ and suppose $Y$ has rational singularities. Then $$\omega_Y^{\bullet}\simeq R\mathscr Hom_Y(\mathscr O_Y, \omega_Y^{\bullet})\simeq R\mathscr Hom_Y(Rf_*\mathscr O_X, \omega_Y^{\bullet})\simeq Rf_*R\mathscr Hom_Y(\mathscr Hom_X(\mathscr O_X, \omega_X^{\bullet})\simeq Rf_*\omega_X[n]\simeq f_*\omega_X[n].$$ (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.

The other direction goes essentially in the same fashion. $\square$

Now try to do this with spectral sequences.

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.)

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## 1 ElementaryEasy

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 $$H^i(Y, f_*\mathscr F)\simeq H^i(X,\mathscr F)$$

Proof Apply the composition rule for the derived functors of $G=\Gamma(Y, \_ )$ and $F=f_*({\_})$. By definition, $G\circ F = \Gamma(X, \_ )$. Then $$R\Gamma(Y, f_*\mathscr F) \simeq R\Gamma(Y, Rf_*\mathscr F) \simeq R\Gamma (X, \mathscr F).$$ Taking cohomology shows the result. $\square$

(edit to please anon, see the comments below) This is usually exhibited as an example of how to use the Leray spectral sequence. Doing it that way is not much harder than the above, but perhaps a bit less "automatic".

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.

This last point is actually an important one regarding the derived category language. You get a higher level notion. The fact that you can work with the complex whose cohomologies give the derived functors of your original functor is very very useful.

## 2 HighSchoolLessEasy

In case you were not convinced by the above example, here is one that should do the trick:

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 $Z$.

Now try to imagine how one could state this using spectral sequences. Both sides actually correspond to spectral sequences, so the statement would be something like "there is a natural map between this an this spectral sequences, such that they converge to the same thing".

I would argue that already the statement of this theorem would be tiring in the language of spectral sequences, but using it would be pure pain.

## 3 CollegeEvenLessEasy

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.

Theorem (a.k.a. Kempf's Criterion) Let $Y$ be a normal variety over $\mathbb C$ with a resolution of singularities $f:X\to Y$. Then $Y$ has rational singularities (i.e., $R^if_*\mathscr O_X=0$ for $i>0$) if and only if

1. $Y$ is Cohen-Macaulay
2. $f_*\omega_X\simeq \omega_Y$

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.

The other direction goes essentially in the same fashion. $\square$

Now try to do this with spectral sequences.

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.)

9 Fixed some typos; [made Community Wiki]

## 1 Elementary

Proposition Let $f:X\to Y$ be a continuous map of topological spaces, $\mathscr F$ a sheaf of abelain 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 $$H^i(Y, f_*\mathscr F)\simeq H^i(X,\mathscr F)$$

Proof Apply the composition rule for the derived functors of $G=\Gamma(Y, \_ )$ and $F=f_*({\_})$. By definition, $G\circ F = \Gamma(X, \_ )$. Then $$R\Gamma(Y, f_*\mathscr F) \simeq R\Gamma(Y, Rf_*\mathscr F) \simeq R\Gamma (X, \mathscr F).$$ Taking cohomology shows the result. $\square$

(edit to please anon, see the comments below) This is usually exhibited as an example of how to use the Leray spectral sequence. Doing it that way is not much harder than the above, but perhaps a bit less "automatic".

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 isomoprhicisomorphic, 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.

This last point is actually an important one regarding the derived category language. You get a higher level notion. The fact that you can work with the complex whose cohomologies give the derived functors of your original functor is very very useful.

## 2 High School

In case you were not convinced by the above example, here is one that should do the trick:

A special case of Grothedndieck 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" dulaizing dualizing complex where $\varepsilon: Z\to \mathrm{Spec}\ k$ is the structure map of $Z$.

Now try to imagine how one could state this using spectral sequences. Both sides actually correspond to spectral sequences, so the statement would be something like "there is a natural map between this an this spectral sequences, such that they converge to the same thing".

I would argue that already the statement of this theorem would be tiring in the language of spectral sequences, but using it would be pure pain.

## 3 College

Here is an application of Grothedieck 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.

Theorem (a.k.a. Kempf's Criterion) Let $Y$ be a normal variety over $\mathbb C$ with a resolution of singularities $f:X\to Y$. Then $Y$ has rational singularities (i.e., $R^if_*\mathscr O_X=0$ for $i>0$) if and only if

1. $Y$ is Cohen-Macaulay
2. $f_*\omega_X\simeq \omega_Y$

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.

The other direction goes essentially in the same fashion. $\square$

Now try to do this with spectral sequences.

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 catgoriescategories, 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.)

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