Lemma 13.9.3 (08RI)—The Stacks project

Processing math: 100%

Table of contents
Part 1: Preliminaries
Chapter 13: Derived Categories
Section 13.9: Cones and termwise split sequences
Lemma 13.9.3 (cite)

numberstags

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

$g$ factors through a morphism $C(f)^\bullet \to M^\bullet$ , and
$f$ 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 September 06, 2013 at 08:24

Just a remark: this lemma is a special case of Lemma 13.8.2 (TAG 014F), namely take $f: K^{\bullet} \to L^{\bullet}$ for $f_1$ , $g: L^{\bullet} \to M^{\bullet}$ for $b$ , and $K^{\bullet}_2 = 0$ .

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).

Comments (1)

Post a comment