Lemma 22.27.8. In Situation 22.27.2 let $x_ i \to y_ i \to z_ i$ be morphisms in $\mathcal{A}$ ($i=1,2,3$) such that $x_2 \to y_2\to z_2$ is an admissible short exact sequence. Let $b : y_1 \to y_2$ and $b' : y_2\to y_3$ be morphisms in $\text{Comp}(\mathcal{A})$ such that

$\vcenter { \xymatrix{ x_1 \ar[d]_0 \ar[r] & y_1 \ar[r] \ar[d]_ b & z_1 \ar[d]_0 \\ x_2 \ar[r] & y_2 \ar[r] & z_2 } } \quad \text{and}\quad \vcenter { \xymatrix{ x_2 \ar[d]^0 \ar[r] & y_2 \ar[r] \ar[d]^{b'} & z_2 \ar[d]^0 \\ x_3 \ar[r] & y_3 \ar[r] & z_3 } }$

commute up to homotopy. Then $b'\circ b$ is homotopic to $0$.

Proof. By Lemma 22.27.5, we can replace $b$ and $b'$ by homotopic maps $\tilde{b}$ and $\tilde{b}'$, such that the right square of the left diagram commutes and the left square of the right diagram commutes. Say $b = \tilde{b} + d(h)$ and $b'=\tilde{b}'+d(h')$ for degree $-1$ morphisms $h$ and $h'$ in $\mathcal{A}$. Hence

$b'b = \tilde{b}'\tilde{b} + d(\tilde{b}'h + h'\tilde{b} + h'd(h))$

since $d(\tilde{b})=d(\tilde{b}')=0$, i.e. $b'b$ is homotopic to $\tilde{b}'\tilde{b}$. We now want to show that $\tilde{b}'\tilde{b}=0$. Because $x_2\xrightarrow {f} y_2\xrightarrow {g} z_2$ is an admissible short exact sequence, there exist degree $0$ morphisms $\pi : y_2 \to x_2$ and $s : z_2 \to y_2$ such that $\text{id}_{y_2} = f\pi + sg$. Therefore

$\tilde{b}'\tilde{b} = \tilde{b}'(f\pi + sg)\tilde{b} = 0$

since $g\tilde{b} = 0$ and $\tilde{b}'f = 0$ as consequences of the two commuting squares. $\square$

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).