Let $(P, d), (P', d')$ be complexes of projective modules, which are zero in all negative degrees. Let $\phi: P\to P'$ be a chain map which induces the trivial morphism on homology in all degrees, $\forall i\; H_i(\phi) = 0.$ Prove that this map is nullhomotopic $\phi\sim 0$.
[Hint: First try to prove the result for the case where $\forall i > 0\; H_i(P) = H_i(P') = 0.]
Thoughts:
$H_i(\phi) = 0\implies \phi_i(\mathrm{im}\; d_{i+1})\subseteq \phi_i(\ker d_i)\subseteq \operatorname{im}d'_{i+1}\subseteq\ker d'_i$.
We want to construct maps
$$h_n: P_n\to P'_{n+1}\quad\phi_n = h_{n-1}\circ d_n + d'_{n+1}\circ h_n$$
In the special case of both complexes acyclic, $H_i(P) = H_i(P') = 0$
$$\phi_i(\mathrm{im}\; d_{i+1})=\phi_i(\ker d_i)\subseteq \operatorname{im}d'_{i+1} = \ker d'_i$$ I'm not sure if this leads anywhere from here.
Best Answer
I will address the case where $P$ and $P'$ are acyclic (as it was noted in a comment that the general case is not true). It actually follows that $\phi$ is null-homotopic from a standard result about projective resolutions (Theorem 2.2.6 in Weibel's book, for example). Since $P$ is acyclic, it is a resolution of the zero module, in the sense that $$ \ldots \to P_2 \to P_1 \to P_0 \to 0 $$ is exact (and similarly for $P'$). Now this standard result says that if I have projective resolutions of two modules (both the zero module in this case) and a map between those two modules (the zero map!)
$$ \require{AMScd} \begin{CD} \ldots @>>> P_2 @>>> P_1 @>>> P_0 @>>> 0 \\ @. @. @. @. @VVV \\ \ldots @>>> P_2 @>>> P_1 @>>> P_0 @>>> 0 \end{CD} $$
then the vertical map lifts to a chain map $P \to P'$ so that the diagram
$$ \require{AMScd} \begin{CD} \ldots @>>> P_2 @>>> P_1 @>>> P_0 @>>> 0 \\ @. @VVV @VVV @VVV @VVV \\ \ldots @>>> P_2 @>>> P_1 @>>> P_0 @>>> 0 \end{CD} $$
commutes. Furthermore, the theorem asserts that this chain map is unique up to chain homotopy, meaning that any two chain maps $P \to P'$ making the diagram above commute must be chain homotopic. Well, your chain map $\phi$ is one such chain map, but so is the zero chain map. Thus $\phi$ and the zero chain map are homotopic.
Let me also indicate how to go about proving the result directly. It's similar to how you prove the theorem I cited, and in general it's the way you approach a statement about bounded below complexes of projectives: by building your maps inductively. First, take your $h_{-1} : 0 \to P_0'$ to be the zero map (your only option). Now we're looking for a map $h_0 : P_0 \to P_1'$ such that $d_1' h_0 = \phi_0$; since $d_1'$ is surjective ($P'$ being acyclic), we immediately get such an $h_0$ from the lifting property of the projective module $P_0$.
For the inductive step, we're looking for $h_n : P_n \to P_{n+1}'$ such that $d_{n+1}' h_n = \phi_n - h_{n-1} d_n$. Now, $\phi_n - h_{n-1} d_n$ lands in $\ker(d_n')$ (check this!), which is the image of $d_{n+1}'$. From here, can you see how to produce $h_n$?