explicit formula for contraction of conus of homotopy equivalence Let $f: K^\bullet \to L^\bullet$ be homotopy equivalence. From theory of triangulated categories it follows that $C(f)^\bullet$ is contractible. But how can i produce explicit formula for homotopy of $Id_{C(f)^\bullet}$ and $0_{C(f)^\bullet}$? 
 A: It is not that difficult! Let us write down carefully all the morphisms and identities that we have. We have morphisms of complexes
$$f^n\colon K^n \to L^n, \quad f^{n+1} \circ d^n_K = d_L^n\circ f^n$$
and
$$g^n\colon L^n \to K^n, \quad g^{n+1} \circ d^n_L = d_K^n\circ g^n,$$
and homotopies $h_K^n\colon K^n\to K^{n-1}$, $h_L^n\colon L^n\to L^{n-1}$ such that
$$\mathrm{id}_{K^{n+1}} - g^{n+1}\circ f^{n+1} = d_K^n\circ h_K^{n+1} + h_K^{n+2}\circ d_K^{n+1}$$
and
$$\mathrm{id}_{L^n} - f^n\circ g^n = d_L^{n-1}\circ h_L^n + h_L^{n+1}\circ d_L^n.$$
The cone $C (f)$ is formed by objects $C (f)^n \mathrel{\mathop:}= L^n\oplus K^{n+1}$ and differentials
$$d^n \mathrel{\mathop:}= \begin{pmatrix}
d^n_L & f^{n+1} \\
0 & -d^{n+1}_K
\end{pmatrix}.$$
We want to construct certain homotopy $h^n\colon C (f)^n \to C (f)^{n-1}$. In theory, it is possible to come up directly with a null-homotopy, i.e. find some $h^n$ such that
$$d^{n-1} \circ h^n + h^{n+1}\circ d^n \stackrel{???}{=} \mathrm{id}_{C (f)^n} = \begin{pmatrix}
\mathrm{id}_{L^n} & 0 \\
0 & \mathrm{id}_{K^{n+1}}
\end{pmatrix},$$
but instead, let us try a rather obvious candidate (which won't be a null-homotopy, but we'll see how to correct it)
$$h^n \mathrel{\mathop:}= \begin{pmatrix}
h^n_L & 0 \\
g^n & -h_K^{n+1}
\end{pmatrix}.$$
Multiplying matrices, we get
$$d^{n-1} \circ h^n + h^{n+1}\circ d^n =$$
$$\begin{pmatrix}
d^{n-1}_L & f^n \\
0 & -d^n_K
\end{pmatrix} \circ \begin{pmatrix}
h^n_L & 0 \\
g^n & -h_K^{n+1}
\end{pmatrix} + \begin{pmatrix}
h^{n+1}_L & 0 \\
g^{n+1} & -h_K^{n+2}
\end{pmatrix} \circ \begin{pmatrix}
d^n_L & f^{n+1} \\
0 & -d^{n+1}_K
\end{pmatrix} =$$
$$\begin{pmatrix}
d_L^{n-1}\circ h_L^n + f^n\circ g^n & -f^n\circ h_K^{n+1} \\
-d_K^n\circ g^n & d^n_K\circ h_K^{n+1}
\end{pmatrix} + \begin{pmatrix}
h_L^{n+1}\circ d_L^n & h_L^{n+1}\circ f^{n+1} \\
g^{n+1}\circ d^n_L & g^{n+1}\circ f^{n+1} + h^{n+2}_K\circ d^{n+1}_K
\end{pmatrix} =$$
$$\begin{pmatrix}
\mathrm{id}_{L^n} & - f^n\circ h_K^{n+1} + h_L^{n+1}\circ f^{n+1} \\
0 & \mathrm{id}_{K^{n+1}}
\end{pmatrix}.$$
Sadly, $k^n \mathrel{\mathop:}= - f^n\circ h_K^{n+1} + h_L^{n+1}\circ f^{n+1} \ne 0$, so $h^n$ is not a null-homotopy but a homotopy between the zero morphism and some morphism $\phi^n \mathrel{\mathop:}= \begin{pmatrix}
\mathrm{id}_{L^n} & k^n \\
0 & \mathrm{id}_{K^{n+1}}
\end{pmatrix}$. However, the latter is visibly an isomorphism of complexes, having $\psi^n \mathrel{\mathop:}= \begin{pmatrix}
\mathrm{id}_{L^n} & -k^n \\
0 & \mathrm{id}_{K^{n+1}}
\end{pmatrix}$ as its inverse. Hence the required null-homotopy is given by $h^n\circ \psi^n$:
$$d^{n-1}\circ h^n\circ \psi^n + h^{n+1}\circ \psi^{n+1}\circ d^n = d^{n-1}\circ h^n\circ \psi^n + h^{n+1}\circ d^n\circ \psi^n = \phi^n\circ \psi^n = \mathrm{id}_{C (f)^n}.$$
By the way, it is also true that if you have a chain contraction for the cone of $f^\bullet$, then $f^\bullet$ is a homotopy (proof: write the chain contraction $2\times 2$ matrix and look at the identity $d^{n-1} \circ h^n + h^{n+1}\circ d^n = \mathrm{id}_{C (f)^n}$).

And to make you laugh, here is this construction as described in a real GTM textbook (Rosenberg, "Algebraic K-Theory and Its Applications", p. 45):


