Lemma 13.9.8. Let $\mathcal{A}$ be an additive category. Let $\alpha : K^\bullet \to L^\bullet $ be a morphism of complexes of $\mathcal{A}$. There exists a factorization

\[ \xymatrix{ K^\bullet \ar[r]^ i \ar@/_1pc/[rr]_\alpha & \tilde K^\bullet \ar[r]^{\tilde\alpha } & L^\bullet } \]

such that

$\tilde\alpha $ is a termwise split surjection (see Definition 13.9.4),

there is a map of complexes $s : \tilde K^\bullet \to K^\bullet $ such that $s \circ i = \text{id}_{K^\bullet }$ and such that $i \circ s$ is homotopic to $\text{id}_{\tilde K^\bullet }$.

Moreover, if both $K^\bullet $ and $L^\bullet $ are in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^ b(\mathcal{A})$, then so is $\tilde K^\bullet $.

**Proof.**
Dual to Lemma 13.9.6. Take

\[ \tilde K^ n = K^ n \oplus L^{n - 1} \oplus L^ n \]

and we define

\[ d^ n_{\tilde K} = \left( \begin{matrix} d^ n_ K
& 0
& 0
\\ 0
& - d^{n - 1}_ L
& \text{id}_{L^ n}
\\ 0
& 0
& d^ n_ L
\end{matrix} \right) \]

in other words $\tilde K^\bullet = K^\bullet \oplus C(1_{L^\bullet [-1]})$. Moreover, we set

\[ \tilde\alpha = \left( \begin{matrix} \alpha
& 0
& \text{id}_{L^ n}
\end{matrix} \right) \]

which is clearly a split surjection. It is also clear that it defines a morphism of complexes. We define

\[ i = \left( \begin{matrix} \text{id}_{K^ n}
\\ 0
\\ 0
\end{matrix} \right) \]

so that clearly $\tilde\alpha \circ i = \alpha $. We set

\[ s = \left( \begin{matrix} \text{id}_{K^ n}
& 0
& 0
\end{matrix} \right) \]

so that $s \circ i = \text{id}_{K^\bullet }$. Finally, let $h^ n : \tilde K^ n \to \tilde K^{n - 1}$ be the map which maps the summand $L^{n - 1}$ of $\tilde K^ n$ via the identity morphism to the summand $L^{n - 1}$ of $\tilde K^{n - 1}$. Then it is a trivial matter to prove that

\[ \text{id}_{\tilde K^\bullet } - i \circ s = d \circ h + h \circ d \]

which finishes the proof of the lemma.
$\square$

## Comments (1)

Comment #293 by arp on