Right homotopic maps iff chain homotopic Assume the model structure on $Ch(R)$ (chain complexes of left modules over the ring $R$) in which fibrations are dimensionwise epimorphisms (i.e. surjections) and weak equivalences are homology isomorphisms (I don't need cofibrations so I won't describe them).
As the title suggests, I need to prove that two chain maps $f,g:B \to X$ are right homotopic $\iff$ they're chain homotopic. The "$\Longleftarrow$" side is easy, since Hovey provides a path object that does the job ($P_n:=X_n\oplus X_n\oplus X_{n+1}$ with $d(x,y,z):=(dx,dy, -dz+x-y))$. 
How can I prove the "$\Longrightarrow$" side, given that the right homotopy could be realized with any path object, not necessarily the one suggested? By the way, I can show it if I assume that the homotopy is indeed obtained by using that particular path object.
Thanks in advance for any hint!!
 A: I tried using the cofibrant replacement functor, since a cofibrant domain simplify the problem (as expressed in the comments).
What I've obtained is that $f \circ q_B \sim g \circ q_B$ (chain homotopic) where $q_B$ is a trivial fibration, since now any path object works and so I can use the one suggested by the author.But unfortunately it doesn't seem to be very helpful: I would like to show that $f,g$ are chain-homotopic themselves, but if $B$ is acyclic, for example, $q_B$ could be the zero map and so not many informations will be grasped.
A: The implication "right homotopic implies chain homotopic" is actually false unless you assume that $B$ is a cofibrant complex.  Here's an example.
Let $B$ be the complex $\Bbb Z/2$ concentrated in degree zero, and let $C$ be the following complex concentrated in (homological) degrees $0$, $1$, $2$:
$$
\cdots 0 \to \Bbb Z \mathop{\longrightarrow}^2 \Bbb Z \mathop{\longrightarrow}^{(1,1)} \Bbb Z/2 \times \Bbb Z/2 \to 0.
$$
We have the two maps $B \to C$ given by the inclusions of the two factors of $\Bbb Z/2$, and they are not chain homotopic (there are no maps from the degree zero component of $B$ to the degree $1$ component of $C$).
I claim that these maps are actually right homotopic, however.  In fact, consider the map $C \to B$ given in degree zero by the map $(n,m) \mapsto n+m$.  This map is a surjection, and if you compute homology of $C$ you find that it is a quasi-isomorphism.  So in fact, we have constructed maps
$$
B \oplus B \to C \mathop{\twoheadrightarrow}^\sim B
$$
where the first map is a cofibration and the second is an acyclic fibration.  This means that $C$ is a path object for $B$, and so any map $C \to Y$ is a right homotopy between the two resulting maps $B \rightrightarrows Y$.  In particular, we may use the identity map $C \to C$ as a right homotopy between the two inclusions.
(As an aside, given a complex $B$, a path object carries a "universal example" of two right homotopic maps, and if they are chain homotopic there then they are chain homotopic everywhere.)
