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The Stacks project

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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