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 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 $K(\mathcal{A})$.
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
(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
(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 $y_ i \to c(f_ i) \to x_ i[1]$.
$\square$
\[ \xymatrix{ x_1 \ar[r]_{f_1} \ar[d]_ a & y_1 \ar[d]^ b \\ x_2 \ar[r]^{f_2} & y_2 } \]
\[ (a, b, c) : (x_1, y_1, c(f_1)) \to (x_1, y_1, c(f_1)) \]
\[ f_ i = \pi _ i \text{d}(s_ i) = - \text{d}(\pi _ i) s_ i \]
\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*}
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (1)
Comment #1040 by Jakob Scholbach on