Lemma 13.9.2 (014F)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 13: Derived Categories
Section 13.9: Cones and termwise split sequences
Lemma 13.9.2 (cite)

\[ \xymatrix{ K_1^\bullet \ar[r]_{f_1} \ar[d]_ a & L_1^\bullet \ar[d]^ b \\ K_2^\bullet \ar[r]^{f_2} & L_2^\bullet } \]

is a diagram of morphisms of complexes which is commutative up to homotopy. Then there exists a morphism $c : C(f_1)^\bullet \to C(f_2)^\bullet $ which gives rise to a morphism of triangles $(a, b, c) : (K_1^\bullet , L_1^\bullet , C(f_1)^\bullet , f_1, i_1, p_1) \to (K_2^\bullet , L_2^\bullet , C(f_2)^\bullet , f_2, i_2, p_2)$ of $K(\mathcal{A})$.

Proof. Let $h^ n : K_1^ n \to L_2^{n - 1}$ be a family of morphisms such that $b \circ f_1 - f_2 \circ a= d \circ h + h \circ d$. Define $c^ n$ by the matrix

\[ c^ n = \left( \begin{matrix} b^ n & h^{n + 1} \\ 0 & a^{n + 1} \end{matrix} \right) : L_1^ n \oplus K_1^{n + 1} \to L_2^ n \oplus K_2^{n + 1} \]

A matrix computation show that $c$ is a morphism of complexes. It is trivial that $c \circ i_1 = i_2 \circ b$, and it is trivial also to check that $p_2 \circ c = a \circ p_1$. $\square$

Comments (3)

Comment #290 by arp on September 06, 2013 at 08:15

Typos:

In the proof, unless I'm misunderstanding your conventions for matrices, I think the matrix for $c$ should be $c^n = \left( \begin{matrix} b^n & h^{n + 1} \\ 0 & a^{n+1} \end{matrix} \right)$

This is silly but I think to really make $c$ a morphism of complexes, you should choose $h$ such that $b \circ f_1 - f_2 \circ a = d \circ h + h \circ d$ , i.e. take $-$ of the $h$ as currently written.

Comment #1554 by jpg on June 29, 2015 at 20:41

I think there is a small typo in the statement of the lemma: the index of the target of the morphism of triangles is 2.

Comment #1572 by Johan on July 14, 2015 at 13:33

Thanks, fixed here.

Comments (3)

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