Homotopic chain maps have homotopy equivalent mapping cones? My question is a general question about the mapping cone of chain maps. Are the mapping cones of homotopic chain maps themselves homotopy equivalent? That is, I want to prove: 

If $f \sim g$ are homotopic chain maps, then the mapping cones $C(f) \simeq C(g)$ are homotopy equivalent complexes.

From a geometric standpoint, it seems like the answer should be yes (e.g. see homotopic maps have homotopy equivalent mapping cones). However, I've tried coming up with chain maps between the complexes $C(f)$ and $C(g)$ without luck. Perhaps there's a better way of seeing it?
 A: Well you're looking for a map $A^{n+1}\oplus B^n\to A^{n+1}\oplus B^n$, and your data is a homotopy $f\to g$, that is, a whole bunch of maps $h_n : A^{n+1}\to B^n$. 
So you might try to use that. 
How about $K:(a,b) \mapsto (a,b+h(a))$ (forgetting the indices) ?
Then let's see how it interacts with the differentials : $d_f(K(a,b)) = d_f(a,b+h(a)) = (d(a), d(b+h(a))+f(a)) = (d(a), d(b)+dh(a) + f(a))$
Now say you have something like $dh+hd = g-f$, so that $dh+f = g-hd$, you get $(d(a), d(b)+g(a) - hd(a))$
Compare with $K(d_g(a,b)) = K((d(a), d(b)+g(a))) = (d(a), d(b)+g(a)+hd(a))$
That's almost the same thing, up to a sign. 
Well this is not a problem : it's just that I most likely messed up some of the sign conventions (for instance, $d$ on $A[1]$ is probably something like $-d_A$ if you pay more attention).
So this is just a rough sketch of what it'll look like, but what I'm saying is that $K$ (or some slight modification of it) will probably be a chain map between $C(g)$ and $C(f)$
Now there will be a similar chain map from $C(f)$ to $C(g)$, and with a bit of luck it'll be easy to show that the two are homotopy inverses to one another.
