Considering each half of factorization of weak equivalence separately

Question

I have been working through the proof of Theorem 3.4.1 in the article https://arxiv.org/pdf/1211.2851.pdf (pages 35-6), but there is one technical detail still unclear to me.

Specifically, we have constructed a commutative diagram

in $\mathbf{sSet}$ such that

1. $A \to B$ is a cofibration,
2. $w$ is a weak equivalence,
3. $E_i\to A$ is a Kan fibration, and
4. $\overline{E}_2 \to B$ is a Kan fibration.

We now want to prove the claim (c) that $\overline{E}_1$ is a fibration over $B$ and $\overline{w}$ is a weak equivalence. The authors state that by factoring the weak equivalence $w$ as a trivial cofibration followed by a trivial fibration, we may prove (c) assuming that $w$ is either a trivial cofibration or a trivial fibration. I'm sure this is obvious to most, but I can't see why this suffices. I would be grateful for an explanation.

Answer 1

Note that the construction $w \mapsto \overline{w}$ is functorial. Thus, if $w = v u$ is a factorization as a trivial cofibration followed by a trivial fibration, we have $\overline{w} = \overline{v} \,\overline{u}$.

Now if (c) holds under the two stated assumptions, the statement of (c) can be applied to $u$ and $v$. Applied to $v$, we see that $\overline{v}$ is a weak equivalence and its domain is a fibration over $B$. The latter conclusion now means that we can apply (c) to $u$, concluding that $\overline{u}$ is a weak equivalence and its domain is a fibration over $B$. But the domain of $\overline{u}$ is $\overline{E_1}$, and $\overline{w} = \overline{v} \,\overline{u}$ so it is also a weak equivalence.