Lemma 22.27.12. In Situation 22.27.2 let $f_1 : x_1 \to y_1$ and $f_2 : x_2 \to y_2$ be morphisms in $\text{Comp}(\mathcal{A})$. Let

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

Lemma 22.27.12. In Situation 22.27.2 let $f_1 : x_1 \to y_1$ and $f_2 : x_2 \to y_2$ be morphisms in $\text{Comp}(\mathcal{A})$. Let

\[ (a,b,c): (x_1,y_1,c(f_1), f_1, i_1, p_1) \to (x_2,y_2, c(f_2), f_2, i_1, p_1) \]

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

**Proof.**
Since $a$ and $b$ are homotopy equivalences, they are invertible in $K(\mathcal{A})$ so let $a^{-1}$ and $b^{-1}$ denote their inverses in $K(\mathcal{A})$, giving us a commutative diagram

\[ \xymatrix{ x_2\ar[d]^{a^{-1}}\ar[r]^{f_2} & y_2\ar[d]^{b^{-1}}\ar[r]^{i_2} & c(f_2)\ar[d]^{c'} \\ x_1\ar[r]^{f_1} & y_1 \ar[r]^{i_1} & c(f_1) } \]

where the map $c'$ is defined via Lemma 22.27.3 applied to the left commutative box of the above diagram. Since the diagram commutes in $K(\mathcal{A})$, it suffices by Lemma 22.27.8 to prove the following: given a morphism of triangle $(1,1,c): (x,y,c(f),f,i,p)\to (x,y,c(f),f,i,p)$ in $K(\mathcal{A})$, the map $c$ is an isomorphism in $K(\mathcal{A})$. We have the commutative diagrams in $K(\mathcal{A})$:

\[ \vcenter { \xymatrix{ y\ar[d]^{1}\ar[r] & c(f)\ar[d]^{c}\ar[r] & x[1]\ar[d]^{1} \\ y\ar[r] & c(f) \ar[r] & x[1] } } \quad \Rightarrow \quad \vcenter { \xymatrix{ y\ar[d]^{0}\ar[r] & c(f)\ar[d]^{c-1}\ar[r] & x[1]\ar[d]^{0} \\ y\ar[r] & c(f) \ar[r] & x[1] } } \]

Since the rows are admissible short exact sequences, we obtain the identity $(c-1)^2 = 0$ by Lemma 22.27.8, from which we conclude that $2-c$ is inverse to $c$ in $K(\mathcal{A})$ so that $c$ is an isomorphism. $\square$

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