The Stacks project

Lemma 13.9.3. Suppose that $f: K^\bullet \to L^\bullet $ and $g : L^\bullet \to M^\bullet $ are morphisms of complexes such that $g \circ f$ is homotopic to zero. Then

  1. $g$ factors through a morphism $C(f)^\bullet \to M^\bullet $, and

  2. $f$ factors through a morphism $K^\bullet \to C(g)^\bullet [-1]$.

Proof. The assumptions say that the diagram

\[ \xymatrix{ K^\bullet \ar[r]_ f \ar[d] & L^\bullet \ar[d]^ g \\ 0 \ar[r] & M^\bullet } \]

commutes up to homotopy. Since the cone on $0 \to M^\bullet $ is $M^\bullet $ the map $C(f)^\bullet \to C(0 \to M^\bullet ) = M^\bullet $ of Lemma 13.9.2 is the map in (1). The cone on $K^\bullet \to 0$ is $K^\bullet [1]$ and applying Lemma 13.9.2 gives a map $K^\bullet [1] \to C(g)^\bullet $. Applying $[-1]$ we obtain the map in (2). $\square$

Comments (1)

Comment #291 by arp on

Just a remark: this lemma is a special case of Lemma 13.8.2 (TAG 014F), namely take for , for , and .

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Your email address will not be published.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 08RI. Beware of the difference between the letter 'O' and the digit '0'.