Decomposition of a fibration into a retraction followed by a minimal fibration My question is about the proof of Theorem 3.5.9. of the book "Model categories" by M. Hovey. The proof is quite long, so I'll try to summarize the facts relevant to this question.
We fix a fibration $p:X\to Y$ and we are considering pairs $(Z,H)$ where $Z$ is a subsimplicial set of  $X$, containing a fixed subsimplicial set $X'$ of $X$ and $H:Z\times \Delta[1]\to X$ is a homotopy from the inclusion $Z\subset X$ to a map $Z\to X'\subset X$ which is constant on $X'\times \Delta[1]$ and covers the constant homotopy on $Z\subset X\to Y$. Such a set can be partially ordered and has a maximal element by Zorn Lemma, let's call it $(Z,H)$. 
We want to prove that $Z$ must coincide with $X$. If not we take an $n$-simplex $x:\Delta[n]\to X$ which is not in $Z$ and such that all the faces of $x$ are in $Z$. We take $Z'$ as the pushout of $\partial\Delta[n]\to \Delta[n]$ with $\partial\Delta[n]\to Z$, the boundary of $x$, and show that we can extend $H$ to a homotopy $H':Z'\times \Delta[1]\to X$ having the required properties, building therefore a couple $(Z',H')$ in contradiction with maximality of $(Z,H)$.
Such an extension is equivalent to a homotopy $H':\Delta[n]\times\Delta[1]\to X$ from $x$ to an $n$-simplex of $X'$, which covers the constant homotopy at $px$ and extends $H$ on $\partial\Delta[n]\times\Delta[1]$.
Now, my doubt is on the equivalent characterization of $H'$ as a $1$-simplex of the fiber of $X^{\Delta[n]}\to X^{\partial\Delta[n]}\times_{Y^{\partial\Delta[n]}}Y^{\Delta[n]}=:P$ over the vertex $(px,H):\Delta[0]\to P$. Indeed, if I'm not mistaken, such a vertex should simply be given by $\partial\Delta[n]\to Z\times \{0\}\to Z\times\Delta[1]\to X$ on $\partial\Delta[n]$, where the last map is the homotopy $H$. In such a case, a $1$-simplex as above would be a homotopy $H':\Delta[n]\times \Delta[1]\to X$ that restricted to $\partial \Delta[n]\times\Delta[1]\to X$ is the constant homotopy at the boundary of $x$, while I don't see why $H$ restricted to $\partial\Delta[n]\times\Delta[1]$ should be constant.
To sum up, my question is

Am I misinterpreting some steps of the proof? Is $H$ constant on $\partial \Delta[n]\times\Delta[1]$. If so, why?

Thank you in advance for any possible help or insight.
 A: As suggested by Andrea in a comment, the proof of Theorem 2.7 in math.univ-toulouse.fr/~dcisinsk/universe.pdf by D.-C. Cisinski is indeed quite similar, and he fixes the mistake in the proof by Hovey. I'll write down a proof in the setting of Theorem 3.5.9, but I'm basically just rephrasing his proof in the context of simplicial sets.
We want to build a homotopy $\Delta[n]\times\Delta[1]\to X$ from $x$ to an $n$-simplex of $X'$ which covers the constant homotopy on $px$ and extends $H$ on $\partial\Delta[n]\times\Delta[1]$.
First we produce a homotopy $H':\Delta[n]\times\Delta[1]\to X$ by considering the map
$$H(\partial x\times 1)\coprod x: \partial\Delta[n]\times\Delta[1] \coprod_{\partial\Delta[n]\times\{0\}}\Delta[n]\times\{0\} \to X$$
Since $p$ is a fibration, we find a lift $H':\Delta[n]\times\Delta[1]\to X$ such that $pH' = px\pi_1$, moreover, we have that $H'$ restricted to $\Delta[n]\times \{0\}$ is equal to $x$, and it extends $H$ on $\partial\Delta[n]\times\Delta[1]$. Now, if we denote by $y_0$ the restriction of $H'$ to $\Delta[n]\times\{1\}$ we have that $y_0$ is not in $X'$ in general. However, $\partial y_0$ is in $X'$ since the restriction of $H$ to $\partial\Delta[n]\times\{1\}$ must factor through $X'$.
Now, $X'$ was built by choosing a set $T$ of simplices of $X$, one for each $p$-equivalence class. Therefore we know that there exist a simplex $y$ in $T$ which lies in the same $p$-equivalence class of $y_0$. Notice that it has been proved earlier in the proof, that a simplex lying on $T$ such that his boundary is a simplex of $X'$, must be itself a simplex of $X'$. Therefore, $y$ lies on $X'$ and what we need to do is to produce a homotopy from $x$ to $y$ with the required properties.
Since $y_0$ and $y$ belong to the same $p$-equivalence class, there exist a homotopy $K:\Delta[n]\times\Delta[1]\to X$ from $y$ to $y_0$ which is constant on $\partial\Delta[n]\times\Delta[1]$ and such that $pK$ is the constant homotopy on $px$.
Using $H'$ and $K$ we can therefore build a map
$$\Delta[n]\times\Lambda^2[2]\coprod_{\partial\Delta[n]\times\Lambda^2[2]}\partial\Delta[n]\times\Delta[2]\to X$$ which is given by $H'$ on $\Delta[n]\times d_1i_2$, by $K$ on $\Delta[n]\times d_0i_2$ and is $H(\partial x\times s_2)$ on $\partial\Delta[n]\times \Delta[2]$.
Solving the lifting problem and taking the restriction to $\Delta[n]\times d_2i_2$ of the lifting $G:\Delta[n]\times\Delta[2]\to X$ we obtain a homotopy from $x$ to $y$ with the required properties.
