Lemma 22.20.6. In Situation 22.20.2 let $\alpha : x \to y$ be a morphism in $\text{Comp}(\mathcal{A})$. Then there exists a factorization in $\text{Comp}(\mathcal{A})$:

\[ \xymatrix{ x \ar[r]^{\tilde{\alpha }} & \tilde{y} \ar@<0.5ex>[r]^{\pi } & y\ar@<0.5ex>[l]^ s } \]

such that

$\tilde{\alpha }$ is an admissible monomorphism, and $\pi \tilde{\alpha }=\alpha $.

There exists a morphism $s:y\to \tilde{y}$ in $\text{Comp}(\mathcal{A})$ such that $\pi s=1_ y$ and $s\pi $ is homotopic to $1_{\tilde{y}}$.

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

\[ \xymatrix@C=3pc{ y \ar@<0.5ex>[r]^ s & y\oplus C(1_ x) \ar@<0.5ex>[l]^{\pi } \ar@<0.5ex>[r]^{p} & C(1_ x)\ar@<0.5ex>[l]^ t } \]

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

\[ \xymatrix{ x \ar[r]^{\tilde{\alpha }} & \tilde{y} \ar@<0.5ex>[r]^{\pi } & y\ar@<0.5ex>[l]^ s } \]

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

\[ \xymatrix{ x \ar@<0.5ex>[r] & C(1_ x) \ar@<0.5ex>[l] \ar@<0.5ex>[r] & x[1]\ar@<0.5ex>[l] } \]

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

\[ \xymatrix{ x\ar[r]^{\tilde{\alpha }} & \tilde{y} \ar[r] & x[1] } \]

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

\begin{align*} 1_{\tilde{y}} - s\pi = & tp \\ = & t(dh)p\quad \text{(by Lemma 09QK)} \\ = & d(thp) \end{align*}

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

## Comments (0)