The Stacks project

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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 013N. Beware of the difference between the letter 'O' and the digit '0'.