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

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

2. 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

Hmm, I don't think $\tilde \alpha$ is a morphism of complexes.

However everything seems OK in the corresponding Lemma 13.8.6 for split injections.

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