The Stacks project

Lemma 13.9.13. Let $\mathcal{A}$ be an additive category. Let $f_1 : K_1^\bullet \to L_1^\bullet $ and $f_2 : K_2^\bullet \to L_2^\bullet $ be morphisms of complexes. Let

\[ (a, b, c) : (K_1^\bullet , L_1^\bullet , C(f_1)^\bullet , f_1, i_1, p_1) \longrightarrow (K_2^\bullet , L_2^\bullet , C(f_2)^\bullet , f_2, i_2, p_2) \]

be any morphism of triangles of $K(\mathcal{A})$. If $a$ and $b$ are homotopy equivalences then so is $c$.

Proof. Let $a^{-1} : K_2^\bullet \to K_1^\bullet $ be a morphism of complexes which is inverse to $a$ in $K(\mathcal{A})$. Let $b^{-1} : L_2^\bullet \to L_1^\bullet $ be a morphism of complexes which is inverse to $b$ in $K(\mathcal{A})$. Let $c' : C(f_2)^\bullet \to C(f_1)^\bullet $ be the morphism from Lemma 13.9.2 applied to $f_1 \circ a^{-1} = b^{-1} \circ f_2$. If we can show that $c \circ c'$ and $c' \circ c$ are isomorphisms in $K(\mathcal{A})$ then we win. Hence it suffices to prove the following: Given a morphism of triangles $(1, 1, c) : (K^\bullet , L^\bullet , C(f)^\bullet , f, i, p)$ in $K(\mathcal{A})$ the morphism $c$ is an isomorphism in $K(\mathcal{A})$. By assumption the two squares in the diagram

\[ \xymatrix{ L^\bullet \ar[r] \ar[d]_1 & C(f)^\bullet \ar[r] \ar[d]_ c & K^\bullet [1] \ar[d]_1 \\ L^\bullet \ar[r] & C(f)^\bullet \ar[r] & K^\bullet [1] } \]

commute up to homotopy. By construction of $C(f)^\bullet $ the rows form termwise split sequences of complexes. Thus we see that $(c - 1)^2 = 0$ in $K(\mathcal{A})$ by Lemma 13.9.12. Hence $c$ is an isomorphism in $K(\mathcal{A})$ with inverse $2 - c$. $\square$

Comments (2)

Comment #1555 by jpg on

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.

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Your email address will not be published.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 014K. Beware of the difference between the letter 'O' and the digit '0'.