Lemma 13.9.6. 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

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

such that

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

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

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

Proof. We set

$\tilde L^ n = L^ n \oplus K^ n \oplus K^{n + 1}$

and we define

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

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

$\tilde\alpha = \left( \begin{matrix} \alpha \\ \text{id}_{K^ n} \\ 0 \end{matrix} \right)$

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

$\pi = \left( \begin{matrix} \text{id}_{L^ n} & 0 & 0 \end{matrix} \right)$

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

$s = \left( \begin{matrix} \text{id}_{L^ n} \\ 0 \\ 0 \end{matrix} \right)$

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

$\text{id}_{\tilde L^\bullet } - s \circ \pi = d \circ h + h \circ d$

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

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).