Lemma 22.27.3. In Situation 22.27.2 suppose that

is a diagram of $\text{Comp}(\mathcal{A})$ commutative up to homotopy. Then there exists a morphism $c : c(f_1) \to c(f_2)$ which gives rise to a morphism of triangles

in $K(\mathcal{A})$.

** The homotopy category is a triangulated category. This lemma proves a part of the axioms of a triangulated category. **

Lemma 22.27.3. In Situation 22.27.2 suppose that

\[ \xymatrix{ x_1 \ar[r]_{f_1} \ar[d]_ a & y_1 \ar[d]^ b \\ x_2 \ar[r]^{f_2} & y_2 } \]

is a diagram of $\text{Comp}(\mathcal{A})$ commutative up to homotopy. Then there exists a morphism $c : c(f_1) \to c(f_2)$ which gives rise to a morphism of triangles

\[ (a, b, c) : (x_1, y_1, c(f_1)) \to (x_1, y_1, c(f_1)) \]

in $K(\mathcal{A})$.

**Proof.**
The assumption means there exists a morphism $h : x_1 \to y_2$ of degree $-1$ such that $\text{d}(h) = b f_1 - f_2 a$. Choose isomorphisms $c(f_ i) = y_ i \oplus x_ i[1]$ of graded objects compatible with the morphisms $y_ i \to c(f_ i) \to x_ i[1]$. Let's denote $a_ i : y_ i \to c(f_ i)$, $b_ i : c(f_ i) \to x_ i[1]$, $s_ i : x_ i[1] \to c(f_ i)$, and $\pi _ i : c(f_ i) \to y_ i$ the given morphisms. Recall that $x_ i[1] \to y_ i[1]$ is given by $\pi _ i \text{d}(s_ i)$. By axiom (C) this means that

\[ f_ i = \pi _ i \text{d}(s_ i) = - \text{d}(\pi _ i) s_ i \]

(we identify $\mathop{\mathrm{Hom}}\nolimits (x_ i, y_ i)$ with $\mathop{\mathrm{Hom}}\nolimits (x_ i[1], y_ i[1])$ using the shift functor $[1]$). Set $c = a_2 b \pi _1 + s_2 a b_1 + a_2hb$. Then, using the equalities found in the proof of Lemma 22.27.1 we obtain

\begin{align*} \text{d}(c) & = a_2 b \text{d}(\pi _1) + \text{d}(s_2) a b_1 + a_2 \text{d}(h) b_1 \\ & = - a_2 b f_1 b_1 + a_2 f_2 a b_1 + a_2 (b f_1 - f_2 a) b_1 \\ & = 0 \end{align*}

(where we have used in particular that $\text{d}(\pi _1) = \text{d}(\pi _1) s_1 b_1 = f_1 b_1$ and $\text{d}(s_2) = a_2 \pi _2 \text{d}(s_2) = a_2 f_2$). Thus $c$ is a degree $0$ morphism $c : c(f_1) \to c(f_2)$ of $\mathcal{A}$ compatible with the given morphisms $y_ i \to c(f_ i) \to x_ i[1]$. $\square$

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## Comments (1)

Comment #1040 by Jakob Scholbach on