Vtyjkyuk

I am aware of this post. The following is a slight generalization,

A free chain complex $(A_*, partial)$ is acyclic iff it has a contracting homotopy.

In that we dont' require $A_n=0$ for $n<0$. Rotman gives a proof as follows.

$partial_n : A_n rightarrow A_n-1$ has image $B_n-1(A_*) = Z_n-1(A_*)$. This induces $s_n-1:Z_n-1(A_*) rightarrow A_n$ s.t. $partial_n s_n-1=1$ as $A_n$ is free.

Now $1-s_n-1 partial_n :A_n rightarrow A_n$ has image in $Z_n(A_*)$. Define $t_n:A_n rightarrow A_n+1$ as the composite,
$$t_n = s_n(1 - s_n-1 partial_n) quad (*)$$
Then $partial_n+1t_n+t_n-1partial_n = 1$.

Induction doesn't really work in this case(?) I would like to hear how one thinks of the formula $(*)$.

I know $s_n, partial_n$ satisfies the retracting property, but its just not obvious.

I do not think one has to use induction: Note that neither the $s_n$, nor the $t_n$ are defined inductively, and the validity of the homotopy property is checked as thus: Obviously $partial_n+1t_n = 1 - s_n-1 partial_n$, and moreover $t_n-1 partial_n = s_n-1 partial_n$ because $partial_n-1 partial_n = 0$.

An alternative proof may be the following: Given a chain complex $C_.$, we may consider the truncated chain complex as follows: Let $n in mathbb Z$ be a whole number, then consider the chain complex
$$
c_n+2 to C_n+1 to C_n to operatornamecoker (delta_n+1) to 0.
$$
This is obviously exact, so we may apply the classical construction to get a chain contraction. But now we can do this for all $n$, and the only thing we have to show is that the definitions coincide to give a chain contraction of the whole complex. (This is probably more complicated than Rotman's proof.)

This is a rephrasing of the formula, but maybe it can help. Consider a general free $mathbb Z$ chain complex $A$. Here is a construction you can do. The surjective maps $partial: A_n to B_n-1(A)$ have splittings $s: B_n-1(A) to A_n$ by freeness, this allows you to break up the chain complex using a map $f: A to Sigma B_*(A) bigoplus Z_*(A)$ given by $a mapsto (partial a, a - spartial a)$. (The chain complex $Sigma B_*(A) bigoplus Z_*(A)$ has differential given by the inclusion $B_n-1(A) to Z_n-1(A)$.) It is easy to check that $f$ is a chain map, and an isomorphism because ignoring differentials it is just a splitting of the exact sequences $0 to Z_n(A) to A_n to B_n-1(A)to 0$.

Assuming that $A_*$ is acyclic now. The contracting homotopy on $Sigma B_*(A) bigoplus Z_*(A)$ is more obvious, it's just given by $h: B_n-1(A) bigoplus Z_n(A) to B_n(A) bigoplus Z_n+1(A)$ where $(b,z) mapsto (z,0)$ using the identification $Z_n(A) = B_n(A)$.