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
$\tilde\alpha $ is a termwise split injection (see Definition 13.9.4),
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.
$\square$
Comments (0)