Lemma 22.9.4. Let $(A, \text{d})$ be a differential graded algebra.

1. Given an admissible short exact sequence $0 \to K \xrightarrow {\alpha } L \to M \to 0$ of differential graded $A$-modules there exists a homotopy equivalence $C(\alpha ) \to M$ such that the diagram

$\xymatrix{ K \ar[r] \ar[d] & L \ar[d] \ar[r] & C(\alpha ) \ar[r]_{-p} \ar[d] & K \ar[d] \\ K \ar[r]^\alpha & L \ar[r]^\beta & M \ar[r]^\delta & K }$

defines an isomorphism of triangles in $K(\text{Mod}_{(A, \text{d})})$.

2. Given a morphism of complexes $f : K \to L$ there exists an isomorphism of triangles

$\xymatrix{ K \ar[r] \ar[d] & \tilde L \ar[d] \ar[r] & M \ar[r]_{\delta } \ar[d] & K \ar[d] \\ K \ar[r] & L \ar[r] & C(f) \ar[r]^{-p} & K }$

where the upper triangle is the triangle associated to a admissible short exact sequence $K \to \tilde L \to M$.

Proof. Proof of (1). We have $C(\alpha ) = L \oplus K$ and we simply define $C(\alpha ) \to M$ via the projection onto $L$ followed by $\beta$. This defines a morphism of differential graded modules because the compositions $K^{n + 1} \to L^{n + 1} \to M^{n + 1}$ are zero. Choose splittings $s : M \to L$ and $\pi : L \to K$ with $\mathop{\mathrm{Ker}}(\pi ) = \mathop{\mathrm{Im}}(s)$ and set $\delta = \pi \circ \text{d}_ L \circ s$ as usual. To get a homotopy inverse we take $M \to C(\alpha )$ given by $(s , -\delta )$. This is compatible with differentials because $\delta ^ n$ can be characterized as the unique map $M^ n \to K^{n + 1}$ such that $\text{d} \circ s^ n - s^{n + 1} \circ \text{d} = \alpha \circ \delta ^ n$, see proof of Homology, Lemma 12.14.10. The composition $M \to C(f) \to M$ is the identity. The composition $C(f) \to M \to C(f)$ is equal to the morphism

$\left( \begin{matrix} s \circ \beta & 0 \\ -\delta \circ \beta & 0 \end{matrix} \right)$

To see that this is homotopic to the identity map use the homotopy $h : C(\alpha ) \to C(\alpha )$ given by the matrix

$\left( \begin{matrix} 0 & 0 \\ \pi & 0 \end{matrix} \right) : C(\alpha ) = L \oplus K \to L \oplus K = C(\alpha )$

It is trivial to verify that

$\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) - \left( \begin{matrix} s \\ -\delta \end{matrix} \right) \left( \begin{matrix} \beta & 0 \end{matrix} \right) = \left( \begin{matrix} \text{d} & \alpha \\ 0 & -\text{d} \end{matrix} \right) \left( \begin{matrix} 0 & 0 \\ \pi & 0 \end{matrix} \right) + \left( \begin{matrix} 0 & 0 \\ \pi & 0 \end{matrix} \right) \left( \begin{matrix} \text{d} & \alpha \\ 0 & -\text{d} \end{matrix} \right)$

To finish the proof of (1) we have to show that the morphisms $-p : C(\alpha ) \to K$ (see Definition 22.6.1) and $C(\alpha ) \to M \to K$ agree up to homotopy. This is clear from the above. Namely, we can use the homotopy inverse $(s, -\delta ) : M \to C(\alpha )$ and check instead that the two maps $M \to K$ agree. And note that $p \circ (s, -\delta ) = -\delta$ as desired.

Proof of (2). We let $\tilde f : K \to \tilde L$, $s : L \to \tilde L$ and $\pi : L \to L$ be as in Lemma 22.7.4. By Lemmas 22.6.2 and 22.9.3 the triangles $(K, L, C(f), i, p)$ and $(K, \tilde L, C(\tilde f), \tilde i, \tilde p)$ are isomorphic. Note that we can compose isomorphisms of triangles. Thus we may replace $L$ by $\tilde L$ and $f$ by $\tilde f$. In other words we may assume that $f$ is an admissible monomorphism. In this case the result follows from part (1). $\square$

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