Grothendieck spectral sequence

Remarks

To prove it, take an injective resolution $X \to I^\bullet$ and a Cartan–Eilenberg resolution of the complex $G(I^\bullet)$. Apply $F$, and compare the two spectral sequences of the resulting double complex. Because $G(I^n)$ is $F$-acyclic, one of them collapses and computes $R^n(F \circ G)(X)$; the other has the stated $E_2$ page. Without the acyclicity hypothesis the derived functor of $F \circ G$ is not computed by this double complex, and the statement can fail. One example is the Leray spectral sequence, where $G = f_*$ for a continuous map $f$ and $F$ is the global sections functor. Another is the Lyndon–Hochschild–Serre spectral sequence, where $G$ and $F$ take invariants under a normal subgroup and under the quotient group. The low degree terms of the spectral sequence give the exact sequence $0 \to R^1 F(G X) \to R^1(F G)(X) \to F(R^1 G(X)) \to R^2 F(G X) \to R^2(F G)(X)$.

Going deeper

For a student

In practice the theorem is used in low degrees, through the five-term exact sequence. It is also used when the $E_2$ page vanishes outside one row or column, so that $R^n(F \circ G)$ can be read off directly. In the language of derived categories it says that $R(F \circ G) \cong RF \circ RG$.