Lemma 13.14.16. Let $\mathcal{A}, \mathcal{B}, \mathcal{C}$ be triangulated categories. Let $S$, resp. $S'$ be a saturated multiplicative system in $\mathcal{A}$, resp. $\mathcal{B}$ compatible with the triangulated structure. Let $F : \mathcal{A} \to \mathcal{B}$ and $G : \mathcal{B} \to \mathcal{C}$ be exact functors. Denote $F' : \mathcal{A} \to (S')^{-1}\mathcal{B}$ the composition of $F$ with the localization functor.

1. If $RF'$, $RG$, $R(G \circ F)$ are everywhere defined, then there is a canonical transformation of functors $t : R(G \circ F) \longrightarrow RG \circ RF'$.

2. If $LF'$, $LG$, $L(G \circ F)$ are everywhere defined, then there is a canonical transformation of functors $t : LG \circ LF' \to L(G \circ F)$.

Proof. In this proof we try to be careful. Hence let us think of the derived functors as the functors

$RF' : S^{-1}\mathcal{A} \to (S')^{-1}\mathcal{B}, \quad R(G \circ F) : S^{-1}\mathcal{A} \to \mathcal{C}, \quad RG : (S')^{-1}\mathcal{B} \to \mathcal{C}.$

Let us denote $Q_ A : \mathcal{A} \to S^{-1}\mathcal{A}$ and $Q_ B : \mathcal{B} \to (S')^{-1}\mathcal{B}$ the localization functors. Then $F' = Q_ B \circ F$. Note that for every object $Y$ of $\mathcal{B}$ there is a canonical map

$G(Y) \longrightarrow RG(Q_ B(Y))$

in other words, there is a transformation of functors $t' : G \to RG \circ Q_ B$. Let $X$ be an object of $\mathcal{A}$. We have

\begin{align*} R(G \circ F)(Q_ A(X)) & = \mathop{\mathrm{colim}}\nolimits _{s : X \to X' \in S} G(F(X')) \\ & \xrightarrow {t'} \mathop{\mathrm{colim}}\nolimits _{s : X \to X' \in S} RG(Q_ B(F(X'))) \\ & = \mathop{\mathrm{colim}}\nolimits _{s : X \to X' \in S} RG(F'(X')) \\ & = RG(\mathop{\mathrm{colim}}\nolimits _{s : X \to X' \in S} F'(X')) \\ & = RG(RF'(X)). \end{align*}

The system $F'(X')$ is essentially constant in the category $(S')^{-1}\mathcal{B}$. Hence we may pull the colimit inside the functor $RG$ in the third equality of the diagram above, see Categories, Lemma 4.22.8 and its proof. We omit the proof this defines a transformation of functors. The case of left derived functors is similar. $\square$

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