The Stacks project

Proof. By axiom (A), we may let $\tilde{y}$ be the differential graded direct sum of $y$ and $C(1_ x)$, i.e., there exists a diagram

where all morphisms are of degree zero, and in $\text{Comp}(\mathcal{A})$. Let $\tilde{y} = y \oplus C(1_ x)$. Then $1_{\tilde{y}} = s\pi + tp$. Consider now the diagram

where $\tilde{\alpha }$ is induced by the morphism $x\xrightarrow {\alpha }y$ and the natural morphism $x\to C(1_ x)$ fitting in the admissible short exact sequence

So the morphism $C(1_ x)\to x$ of degree 0 in this diagram, together with the zero morphism $y\to x$, induces a degree-0 morphism $\beta : \tilde{y} \to x$. Then $\tilde{\alpha }$ is an admissible monomorphism since it fits into the admissible short exact sequence

Furthermore, $\pi \tilde{\alpha } = \alpha$ by the construction of $\tilde{\alpha }$, and $\pi s = 1_ y$ by the first diagram. It remains to show that $s\pi$ is homotopic to $1_{\tilde{y}}$. Write $1_ x$ as $d(h)$ for some degree $-1$ map. Then, our last statement follows from

since $dt = dp = 0$, and $t$ is of degree zero. $\square$