# 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.

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 $$[f]=[f']$$.