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I heard many times that because of introducing derived category, we can avoid cumbersome spectral sequence. However, I don't quite understand its meaning. Here is a precise example people talking about:

Let $f: X \to Y, g: Y \to Z$ be morphism of smooth varieties, and Let $D(X), D(Y), D(Z)$ be derived category of bounded complexes of quasi-coherent sheaves. Then it is said that because of the associativity of derived functor, i.e.

$${\rm{R}}f_* \circ {\rm{R}}g_* = {\rm{R}}{(f \circ g)}_*\quad,$$

we can avoid the write the spectral sequence.

I don't know which spectral sequence people refer to, and why the associativity of derived functor equivalent to that spectral sequence?

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The spectral sequence is the Leray spectral sequence $$E_2^{i,j} := R^if_* R^jg_* \mathcal F\implies R^{i+j}(f\circ g)_*\mathcal F.$$

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  • $\begingroup$ Ahaaa, I see, thank you very much! Besides, should that spectral sequence be called Grothendieck spectral sequence? $\endgroup$ – Li Yutong May 3 '14 at 13:47
  • $\begingroup$ @LiYutong: Dear Li Yutong, Maybe. (If $f$ is the map to a point, then it is the Leray spectral sequence.) Regards, $\endgroup$ – Matt E May 3 '14 at 15:14
  • $\begingroup$ Dear Matt, another thing I forgot to ask is how can we use the associativity of derived functor to get this spectral sequence (i.e. without using spectral sequence associated to the double complex)? $\endgroup$ – Li Yutong May 3 '14 at 18:58
  • $\begingroup$ Dear Li Yutong, You do have to use the spectral sequence associated to a double complex. (Maybe there is another way, but the spectral sequence of a double complex is what I would use.) At least for me, the point is that the language of derived categories allows us to talk about $Rf_*$, $Rg_*$, etc., without having to unpack everything in terms of complexes, double complexes, and spectral sequences. Regards, $\endgroup$ – Matt E May 4 '14 at 0:25

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