## 13.22 Composition of right derived functors

Sometimes we can compute the right derived functor of a composition. Suppose that $\mathcal{A}, \mathcal{B}, \mathcal{C}$ be abelian categories. Let $F : \mathcal{A} \to \mathcal{B}$ and $G : \mathcal{B} \to \mathcal{C}$ be left exact functors. Assume that the right derived functors $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$, $RG : D^{+}(\mathcal{B}) \to D^{+}(\mathcal{C})$, and $R(G \circ F) : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{C})$ are everywhere defined. Then there exists a canonical transformation

\[ t : R(G \circ F) \longrightarrow RG \circ RF \]

of functors from $D^{+}(\mathcal{A})$ to $D^{+}(\mathcal{C})$, see Lemma 13.14.16. This transformation need not always be an isomorphism.

Lemma 13.22.1. Let $\mathcal{A}, \mathcal{B}, \mathcal{C}$ be abelian categories. Let $F : \mathcal{A} \to \mathcal{B}$ and $G : \mathcal{B} \to \mathcal{C}$ be left exact functors. Assume $\mathcal{A}$, $\mathcal{B}$ have enough injectives. The following are equivalent

$F(I)$ is right acyclic for $G$ for each injective object $I$ of $\mathcal{A}$, and

the canonical map

\[ t : R(G \circ F) \longrightarrow RG \circ RF. \]

is isomorphism of functors from $D^{+}(\mathcal{A})$ to $D^{+}(\mathcal{C})$.

**Proof.**
If (2) holds, then (1) follows by evaluating the isomorphism $t$ on $RF(I) = F(I)$. Conversely, assume (1) holds. Let $A^\bullet $ be a bounded below complex of $\mathcal{A}$. Choose an injective resolution $A^\bullet \to I^\bullet $. The map $t$ is given (see proof of Lemma 13.14.16) by the maps

\[ R(G \circ F)(A^\bullet ) = (G \circ F)(I^\bullet ) = G(F(I^\bullet ))) \to RG(F(I^\bullet )) = RG(RF(A^\bullet )) \]

where the arrow is an isomorphism by Lemma 13.16.7.
$\square$

Lemma 13.22.2 (Grothendieck spectral sequence). With assumptions as in Lemma 13.22.1 and assuming the equivalent conditions (1) and (2) hold. Let $X$ be an object of $D^{+}(\mathcal{A})$. There exists a spectral sequence $(E_ r, d_ r)_{r \geq 0}$ consisting of bigraded objects $E_ r$ of $\mathcal{C}$ with $d_ r$ of bidegree $(r, - r + 1)$ and with

\[ E_2^{p, q} = R^ pG(H^ q(RF(X))) \]

Moreover, this spectral sequence is bounded, converges to $H^*(R(G \circ F)(X))$, and induces a finite filtration on each $H^ n(R(G \circ F)(X))$.

For an object $A$ of $\mathcal{A}$ we get $E_2^{p, q} = R^ pG(R^ qF(A))$ converging to $R^{p + q}(G \circ F)(A)$.

**Proof.**
We may represent $X$ by a bounded below complex $A^\bullet $. Choose an injective resolution $A^\bullet \to I^\bullet $. Choose a Cartan-Eilenberg resolution $F(I^\bullet ) \to I^{\bullet , \bullet }$ using Lemma 13.21.2. Apply the second spectral sequence of Lemma 13.21.3.
$\square$

## Comments (0)