Hello I am trying to prove this
" Given a projective resolution of projectives
$P: \cdots\rightarrow P_{2}\rightarrow P_{1}\rightarrow P_{0} \rightarrow 0$
then $P$ is acyclic; i.e. $H_{n}(P)=0$ for $n>0$, if and only if the identity chain map $Id_{P}$ is a contracting map"
I proved the direction $(\Longleftarrow)$ however for the another direction I have the chain complex
$ \cdots\rightarrow P_{2}\rightarrow P_{1}\rightarrow P_{0} \rightarrow H_{0}(P) \rightarrow 0$
But I do not know how to start with the homotopy maps $S_{0}:P_{0}\rightarrow P_{1}$ in this case.
I only have that for the identity $1:H_{0}(P)\rightarrow H_{0}(P)$ we have chain map $f:P\rightarrow P$ inducing $1$ and it is unique up to homotopy.
From this I have that $f$ is homotopic to $Id_{P}$
Thank you for your time!
---------------Edit-------------
This one idea
Suppose that $H(P) = 0$. Then the maps $1$ and $0 : P \rightarrow P$ induce the same map – namely $φ = 0$ - on $H(P)$. By the comparison theorem we have that the two maps are homotopic
Is my argument correct?