Lemma 22.9.3. Let $(A, \text{d})$ be a differential graded algebra. Let $f_1 : K_1 \to L_1$ and $f_2 : K_2 \to L_2$ be homomorphisms of differential graded $A$-modules. Let

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

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

**Proof.**
Let $a^{-1} : K_2 \to K_1$ be a homomorphism of differential graded $A$-modules which is inverse to $a$ in $K(\text{Mod}_{(A, \text{d})})$. Let $b^{-1} : L_2 \to L_1$ be a homomorphism of differential graded $A$-modules which is inverse to $b$ in $K(\text{Mod}_{(A, \text{d})})$. Let $c' : C(f_2) \to C(f_1)$ be the morphism from Lemma 22.6.2 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 $K(\text{Mod}_{(A, \text{d})})$ then we win. Hence it suffices to prove the following: Given a morphism of triangles $(1, 1, c) : (K, L, C(f), f, i, p)$ in $K(\text{Mod}_{(A, \text{d})})$ the morphism $c$ is an isomorphism in $K(\text{Mod}_{(A, \text{d})})$. By assumption the two squares in the diagram

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

commute up to homotopy. By construction of $C(f)$ the rows form admissible short exact sequences. Thus we see that $(c - 1)^2 = 0$ in $K(\text{Mod}_{(A, \text{d})})$ by Lemma 22.7.6. Hence $c$ is an isomorphism in $K(\text{Mod}_{(A, \text{d})})$ with inverse $2 - c$.
$\square$

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