Lemma 22.20.1. Let $\mathcal{A}$ be a differential graded category satisfying axioms (A) and (B). Given an admissible short exact sequence $x \to y \to z$ we obtain (see proof) a triangle

$x \to y \to z \to x[1]$

in $\text{Comp}(\mathcal{A})$ with the property that any two compositions in $z[-1] \to x \to y \to z \to x[1]$ are zero in $K(\mathcal{A})$.

Proof. Choose a diagram

$\xymatrix{ x \ar[rr]_1 \ar[rd]_ a & & x \\ & y \ar[ru]_\pi \ar[rd]^ b & \\ z \ar[rr]^1 \ar[ru]^ s & & z }$

giving the isomorphism of graded objects $y \cong x \oplus z$ as in the definition of an admissible short exact sequence. Here are some equations that hold in this situation

1. $1 = \pi a$ and hence $\text{d}(\pi ) a = 0$,

2. $1 = b s$ and hence $b \text{d}(s) = 0$,

3. $1 = a \pi + s b$ and hence $a \text{d}(\pi ) + \text{d}(s) b = 0$,

4. $\pi s = 0$ and hence $\text{d}(\pi )s + \pi \text{d}(s) = 0$,

5. $\text{d}(s) = a \pi \text{d}(s)$ because $\text{d}(s) = (a \pi + s b)\text{d}(s)$ and $b\text{d}(s) = 0$,

6. $\text{d}(\pi ) = \text{d}(\pi ) s b$ because $\text{d}(\pi ) = \text{d}(\pi )(a \pi + s b)$ and $\text{d}(\pi )a = 0$,

7. $\text{d}(\pi \text{d}(s)) = 0$ because if we postcompose it with the monomorphism $a$ we get $\text{d}(a\pi \text{d}(s)) = \text{d}(\text{d}(s)) = 0$, and

8. $\text{d}(\text{d}(\pi )s) = 0$ as by (4) it is the negative of $\text{d}(\pi \text{d}(s))$ which is $0$ by (7).

We've used repeatedly that $\text{d}(a) = 0$, $\text{d}(b) = 0$, and that $\text{d}(1) = 0$. By (7) we see that

$\delta = \pi \text{d}(s) = - \text{d}(\pi ) s : z \to x[1]$

is a morphism in $\text{Comp}(\mathcal{A})$. By (5) we see that the composition $a \delta = a \pi \text{d}(s) = \text{d}(s)$ is homotopic to zero. By (6) we see that the composition $\delta b = - \text{d}(\pi )sb = \text{d}(-\pi )$ is homotopic to zero. $\square$

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