Model categories: Any two solutions to a lifting problem are homotopic I am reading the paper "Model Categories and Simplicial Methods" by Goerss and Schemmerhorn  and I am struggling to prove Lemma 2.11, which states the following.

Suppose that there is a lifting problem in a model category $\mathcal{C}$ (shown below), where $j$ is a cofibration, $q$ is a fibration, and one of $j$ or $q$ is a weak equivalence. Then $f$ exists and is unique up to left homotopy under $A$ and over $Y$.




The paper says that this Lemma follows from the model category axioms, but I don't see how.
My attempt
I don't have any very promising angles, but I will list a few things that I have tried.
By the model category axioms, the lifting problem admits a solution (i.e. at least one $f$ exists), so the difficulty is proving that for any two such solutions $f, f'$, we have a (left) homotopy under $A$ and over $Y$ from $f$ to $f'$. I am not very confident with the notions of homotopies over and under objects, so I have just been trying to prove that there is a left homotopy from $f$ to $f'$.
Any homotopy is a factoring of the map
$$
f\coprod f':B\coprod B \to X.
$$
The model category axioms tell us that there is a factorisation
$$
A\coprod A \overset{\alpha}{\to} Z \overset{\beta}{\to} X,
$$
where $\alpha$ is a cofibration and $\beta$ is an acyclic fibration. This factorisation looks a bit like a homotopy, particularly because $i$ is a cofibration. Therefore it would suffice for $Z$ to be a cylinder object of $A$. However, I do not see any reason why $Z$ should be a cylinder object, so this seems to be a dead end.
 A: Suppose that we have two lifts $f,f'$ $q$ is a weak equivalence (the two cases are dual), so we hava a diagram as below 
Since this diagrams commutes, we have a lift $\eta$ which is a left homotopy between  $f$ and $f'$.  Also, if all these objects were fibrant and cofibrant (which most of the time we require them to be), we would have that $f \circ j = f' \circ j$,  so that $[f \circ j]=[f] \circ [j] = [f'] \circ [j]$, and since $j$ is a weak equivalence, [j] is an isomorphism, and so we could've canceled out $[j]$ and arrive at $[f]=[f']$.
A: I'm not sure I understand your claim that the two cases are dual. If $j$ is not acyclic in your diagram, then the lift $\eta$ isn't required to exist. If anything, the dual statement is that $f$ and $f'$ are right homotopic if $q$ is a weak equivalence.
Unless $B$ is cofibrant and $X$ is fibrant, this doesn't imply a left homotopy between the maps. I feel that the article Goerss and Schemerhorn is being a bit sloppy in their statement here. I've not been able to find this statement in this general form anywhere else and it seems quite strong. If I'm wrong here please do let me know!
