Lemma 13.4.22. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. Let $G : \mathcal{A} \to \mathcal{D}$ be a $\delta$-functor. Let $H : \mathcal{D} \to \mathcal{B}$ be a homological functor. Assume that $H^{-1}(G(A)) = 0$ for all $A$ in $\mathcal{A}$. Then the collection

$\{ H^ n \circ G, H^ n(\delta _{A \to B \to C})\} _{n \geq 0}$

is a $\delta$-functor from $\mathcal{A} \to \mathcal{B}$, see Homology, Definition 12.12.1.

Proof. The notation signifies the following. If $0 \to A \xrightarrow {a} B \xrightarrow {b} C \to 0$ is a short exact sequence in $\mathcal{A}$, then

$\delta = \delta _{A \to B \to C} : G(C) \to G(A)[1]$

is a morphism in $\mathcal{D}$ such that $(G(A), G(B), G(C), a, b, \delta )$ is a distinguished triangle, see Definition 13.3.6. Then $H^ n(\delta ) : H^ n(G(C)) \to H^ n(G(A)[1]) = H^{n + 1}(G(A))$ is clearly functorial in the short exact sequence. Finally, the long exact cohomology sequence (13.3.5.1) combined with the vanishing of $H^{-1}(G(C))$ gives a long exact sequence

$0 \to H^0(G(A)) \to H^0(G(B)) \to H^0(G(C)) \xrightarrow {H^0(\delta )} H^1(G(A)) \to \ldots$

in $\mathcal{B}$ as desired. $\square$

Comment #8844 by Z. He on

I think $G$ should be additive, since Definition 12.12.1 requires the cohomological $\delta$-functors to be additive.

Comment #9241 by on

OK, I think it is easy to see that a $\delta$-functor as in Definition 13.3.6 is additive, but I am totally willing to add that as a condition to the definition if people feel this is best.

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