Lemma 13.6.11. Let $H : \mathcal{D} \to \mathcal{A}$ be a homological functor from a triangulated category $\mathcal{D}$ to an abelian category $\mathcal{A}$, see Definition 13.3.5. The subcategory $\mathop{\mathrm{Ker}}(H)$ of $\mathcal{D}$ is a strictly full saturated triangulated subcategory of $\mathcal{D}$ whose corresponding saturated multiplicative system (see Lemma 13.6.10) is the set

\[ S = \{ f \in \text{Arrows}(\mathcal{D}) \mid H^ i(f)\text{ is an isomorphism for all }i \in \mathbf{Z}\} . \]

The functor $H$ factors through the quotient functor $Q : \mathcal{D} \to \mathcal{D}/\mathop{\mathrm{Ker}}(H)$.

**Proof.**
The category $\mathop{\mathrm{Ker}}(H)$ is a strictly full saturated triangulated subcategory of $\mathcal{D}$ by Lemma 13.6.3. The set $S$ is a saturated multiplicative system compatible with the triangulated structure by Lemma 13.5.5. Recall that the multiplicative system corresponding to $\mathop{\mathrm{Ker}}(H)$ is the set

\[ \left\{ \begin{matrix} f \in \text{Arrows}(\mathcal{D}) \text{ such that there exists a distinguished triangle }
\\ (X, Y, Z, f, g, h)\text{ with } H^ i(Z) = 0 \text{ for all }i
\end{matrix} \right\} \]

By the long exact cohomology sequence, see (13.3.5.1), it is clear that $f$ is an element of this set if and only if $f$ is an element of $S$. Finally, the factorization of $H$ through $Q$ is a consequence of Lemma 13.6.8.
$\square$

## Comments (2)

Comment #333 by arp on

Comment #335 by Johan on

There are also: