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13.25 Right derived functors via resolution functors

The content of the following lemma is that we can simply define $RF(K^\bullet ) = F(j(K^\bullet ))$ if we are given a resolution functor $j$.

Lemma 13.25.1. Let $\mathcal{A}$ be an abelian category with enough injectives Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor into an abelian category. Let $(i, j)$ be a resolution functor, see Definition 13.23.2. The right derived functor $RF$ of $F$ fits into the following $2$-commutative diagram

\[ \xymatrix{ D^{+}(\mathcal{A}) \ar[rd]_{RF} \ar[rr]^{j'} & & K^{+}(\mathcal{I}) \ar[ld]^ F \\ & D^{+}(\mathcal{B}) } \]

where $j'$ is the functor from Lemma 13.23.6.

Proof. By Lemma 13.20.1 we have $RF(K^\bullet ) = F(j(K^\bullet ))$. $\square$

Remark 13.25.2. In the situation of Lemma 13.25.1 we see that we have actually lifted the right derived functor to an exact functor $F \circ j' : D^{+}(\mathcal{A}) \to K^{+}(\mathcal{B})$. It is occasionally useful to use such a factorization.

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