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 $2 - c$. $\square$

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.

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).