Let's consider the category $Ch_R$ of cochain complexes of modules over a commutative ring $R$. I'm trying to prove that if the chain map $\phi:M\rightarrow N$ is a homotopy equivalence then its mapping cone is contractible, WITHOUT using results from triangulated categories (i.e. by crude computation): by hypothesis i have that there is $\psi:N\rightarrow M$ such that
$$ \psi^k\phi^k-1^k_M=d^{k-1}_M\rho^k+\rho^{k+1}d_M^k \quad and\quad \phi^k\psi^k-1^k_N=d^{k-1}_N\sigma^k+\sigma^{k+1}d_N^k $$ for some $\rho:M\rightarrow M[-1]$ and $\sigma:N\rightarrow N[-1]$, and i need to find $R:cone\phi\rightarrow cone\phi[-1]$ such that $$ 1^k_{cone\phi}=\bigg( \matrix{1^k_{M[1]} & 0 \\ 0 & 1^k_N} \bigg)=\bigg(\matrix{-d^k_M & 0 \\ \phi^k & d_N^{k-1}} \bigg)R^k+R^{k+1}\bigg(\matrix{-d_M^{k+1} & 0 \\ \phi^{k+1} & d_N^k} \bigg). $$ I already worked out that $R^k=\bigg(\matrix{\rho^{k+1} & \psi^k \\ X^k & -\sigma^k}\bigg)$, but i'm stuck with this last equation
$$ \sigma^{k+1}\phi^{k+1}-\phi^k\rho^{k+1}=d^{k-1}_NX^k-X^{k+1}d^{k+1}_M\quad(*) $$
which is supposed to say that $(\sigma\phi)[1],\phi\rho[1]:M[1]\rightarrow N$ are chain homotopic (or at least they would be if they were chain maps), or that $\phi$ commutes with the chain homotopies if $X$ is a closed morphism of degree -2 (i would like to set $X=0$, but that seems unlikely).
Does this make sense to anyone?
EDIT: If conversely $cone\phi$ is contractible then there is a morphism
$$ R=\bigg(\matrix{\rho[1] & \psi \\ X & -\sigma}\bigg):cone\phi\rightarrow cone\phi[-1] $$
such that $1_{cone\phi}=d_{cone\phi}[-1]R+R[1]d_{cone\phi}$, and thus $\psi$ is a homotopy inverse of $\phi$ and all the conditions above are satisfied.
Since I'm pretty sure that being a homotopy equivalence and having a contractible cone are equivalent conditions there must be a way to prove that equation (*) always holds (I think it all comes down to showing that the square
$$ \begin{array}{ccc} M & \stackrel{\rho}{\longrightarrow} & M[-1] \\ \downarrow{\phi} & & \downarrow{\phi[-1]} \\ N & \stackrel{\sigma}{\longrightarrow} & N[-1] \end{array} $$
is homotopy commutative, at least for a suitable choice of $\rho$ and $\sigma$).
EDIT2: I think that the following diagram is homotopy commutative, and should solve the problem
$$ \begin{array}{ccc} M & \stackrel{\rho}{\longrightarrow} & M[-1] \\ \downarrow{\phi} & \approx & \downarrow{\phi[-1]} \\ N & \stackrel{\phi[-1]\rho\psi}{\longrightarrow} & N[-1] \\ \downarrow{1} & \approx & \downarrow{1} \\ N & \stackrel{\sigma}{\longrightarrow} & N[-1] \end{array} $$
But I still would like someone else's opinion about it.
Since I ran into this problem myself and this question is one of the first results via google, it seems helpful to leave an answer here:
The explicit construction is given in the proof of Proposition 1.1 in Ranicki, "The algebraic theory of finiteness obstruction". The desired chain homotopy is given by \begin{pmatrix} k-fhg+kfg & (-1)^{r+1}fhh-kfh \\ (-1)^{r-1}g & h \\ \end{pmatrix} in Ranickis notation. Ranicki explains his notation right before the Proposition. Roughly, $f=\phi$, $g=\psi$, $h=\rho$ and $k=\sigma$, $r=n$ and the summands of $cone(f)$ are flipped.
The calculation to show that this defines a contraction (which Ranicki omits) works by plugging in the fundamental equations of the chain homotopies ($ch+hc=Id-gf$ and $dk+kd=Id-fg$) at every opportunity.