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.
$\square$

## Comments (0)

There are also: