Let $\mathscr{C}$ be an abelian category. If $P_\bullet \in \operatorname{Ch}_{\geq 0}(\mathscr{C})$ is a bounded below complex of projectives, and $C_\bullet \in \operatorname{Ch}_{\geq 0}(\mathscr{C})$ is a bounded below exact complex, then $[P_\bullet, C_\bullet] = 0$. (Every chain map $P_\bullet \to C_\bullet$ is nullhomotopic.)
It is tempting to conjecture that
If $\mathscr{B} \underset{G}{\overset{F}{\rightrightarrows}} \operatorname{Ch}_{\geq 0}(\mathscr{C})$ are functors from an arbitrary category $\mathscr{B}$, where $F$ lands in projective complexes, and $G$ lands in acyclic complexes, then $[F,G]=0$, meaning that any natural transformation from $F \Rightarrow G$ is naturally chain homotopic to the zero natural transformation.
The acyclic models theorem implies something similar: that if $\mathscr{B}$ has models $\mathcal{M}$, and $F$ is a free functor w.r.t. $\mathcal{M}$, and if $G$ is acyclic, then $[F,G]$ is indeed zero.
Is the highlighted theorem above untrue?
Given a natural transformation, I can choose a map $\tau: (FX)_\bullet \to (GX)_\bullet[1]$ for each object $X \in \mathscr{B}$. Making $\tau$ natural is the problem. If I try to define the first map $\tau_0$ in the natural transformation $\tau$, and check whether it is natural, I find that the naturality diagram \begin{array}{} FX_0 & \xrightarrow{(Ff)_0} &FY_0\\ (\tau_X)_0 \downarrow && \downarrow (\tau_Y)_0 \\ GX_1 & \xrightarrow{(Gf)_1} & GY_1 \end{array} only commutes up to a boundary element in $\partial^{GY}(GY_2)$.
The trouble is that an object of the functor category $[\cal B,\cal C]$ need not be projective even if every object in the image is. I think what you would get would be a weak homotopy (homotopy at each object of $\cal B$). For more on this subject, we my book titled Acyclic Models, which deals with every version of the theorem I am aware of.