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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 $Y$.

enter image description here

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.

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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 enter image description here

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 $[f]=[f']$.

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