Let $f:A\rightarrow B$ be a chain map, i.e. $A,B$ are chain complexes. The mapping cone of $f$, denoted by $C(f)$, is the chain complex with $C(f)_n=A_{n-1}\oplus B_n$ and differential $D(a,b)=(-da,db-f(a)).$
I want to develop a criterion for $f$ being null-homotopic, i.e. $f$ beging chain-homotopic to the zero (chain-)map in terms of $C(f)$.
I started to show that $C(f)$ is indeed a chain complex. That is done now. But for the main part, I don't get along well.
Best Answer
Let $f:A\to B$ be a chain map. Then, applying the cone construction you have above to the identity chain map $id_A:A\to A$ gives a complex $CA = C(id_A)$ with $CA_n = A_{n-1}\oplus A_n$ and differentials $$d:CA_n\to CA_{n-1},\quad d(x,y) = (-dx,dy-x).$$ What I meant by an extension in the comment above is that there exists a chain map $\bar{f}:CA\to B$ such that $$\bar{f_n}(0,x) = f_n(x),\quad \forall x\in A_n.$$ We then have the following result:
For the direction $(\Rightarrow)$, if $f$ is null-homotopic, there exists a sequence of maps $$\{s_n:A_n\to B_{n+1}\}$$ such that $f_n = d_{n+1}^B\circ s_n+s_{n-1}\circ d_n^A$ for all $n$. Then, we define an extension $\bar{f}:CA\to B$ by: $$\bar{f_n}:CA_n\to B_n,\quad \bar{f_n}(x,y) = f_n(y)-s_{n-1}(x).$$ Showing that this is a chain map is elementary, so I'll leave it as an exercise.
For the direction $(\Leftarrow)$, you can use the condition that $\bar{f}$ is a chain map to build chain homotopies: $$s_{n-1}:A_{n-1}\to B_n,\quad s_{n-1}(a) = -\bar{f_n}(a,0).$$ There's nothing fancy needed to show this gives a chain homotopy, so I'll leave this as an exercise as well.