## 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

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

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

threefunctors. On the other hand, when you want to actually compute something, spectral sequences are your only friend, mostly. $\endgroup$ – Mariano Suárez-Álvarez Mar 21 '12 at 2:50