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.

  • $\begingroup$ There is a criterion in terms of the cone $C(id_A)=CA$; A chain map $f:A\to B$ is null homotopic if and only if it extends to a map on the cone $\bar{f}:CA\to B$. I can elaborate on this if you'd like. $\endgroup$ – user171308 Apr 10 '17 at 1:44
  • $\begingroup$ Yes, that would be great. What do you mean with "extends" here? $\endgroup$ – user372565 Apr 10 '17 at 7:18

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:

A chain map $f:A\to B$ is null-homotopic if and only if there exists an extension $\bar{f}:CA\to B$ of $f$ to the cone $CA$.

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.

  • $\begingroup$ Thanks a lot! You mean $\bar{f}_n(0,x)$ instead of $\bar{f}(x,0)$, don't you? $\endgroup$ – user372565 Apr 11 '17 at 13:38
  • $\begingroup$ @user372565 Happy to help! And yes you are exactly right, I've edited a couple typos. $\endgroup$ – user171308 Apr 11 '17 at 17:09
  • $\begingroup$ In the last expression, would you happen to mean to say $- \bar{f}_n(a, 0)$ instead? Otherwise the chain homotopy doesn't work. I would have proposed an edit, but SE requires at least 6 characters for an edit :P $\endgroup$ – Elliot Yu Jul 28 '17 at 2:52
  • $\begingroup$ @Elliot Yu Yes, I think you're right, since $a$ should be in $A_{n-1}$ and not $A_n$. Thanks for letting me know! $\endgroup$ – user171308 Jul 28 '17 at 18:18

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