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