Proof of cellular approximation theorem I'd like to prove the cellular approximation theorem based on the proof given on Tom Dieck (p.$211$, Theorem $8.5.4$).

The proof can be found here: https://www.maths.ed.ac.uk/~v1ranick/papers/diecktop.pdf (p.$211$, Theorem $8.5.4$).
As stated in the book, we want to construct homotopies as follows :

There are some facts in the proof of the inductive step that I'm unable to solve. The proof of the inductive step proceeds as follows:
Without loss of generality we can assume $f(X^i) \subset Y^i$ for $i<n$. Let $\phi_\alpha: (E_{\alpha}^n,S_{\alpha}^{n-1})\longmapsto (X,X^{n-1})$ the attaching map of an $n-$cell (so the image of $\phi_{\alpha}$ lies in $X^{n}$), and $f \circ \phi_{\alpha}:(E_{\alpha}^n,S_{\alpha}^{n-1}) \longmapsto (Y,Y^{n-1})\subset  (Y^{n},Y^{n-1})$.
Since $\pi_n(Y,Y^n) = 0$ we have that $f \circ \phi_{\alpha}$ is homotopic (with homotopy costant on $S_{\alpha}^{n-1}$) to $g_{\alpha}: (E_{\alpha}^n,S_{\alpha}^{n-1}) \longmapsto (Y^n,Y^{n-1})$.
We can repeat this process for every $n-$cell and so we have an homotopy $H^n$ between $f_{|_{X^{n}}}$ and a map $X^n \longmapsto Y^n$.
Since $X^n \hookrightarrow X$ is a cofibration we can extend the homotopy $H^n$ between $f$ and $\tilde{f} : X \longmapsto Y$ such that $\tilde{f}_{|_{X^{n}}} \subset Y^n$ and $H^n$ costant on $X^{n-1}$.
$\rule{18cm}{1pt}$
Edit $1$ : The points I'd like to clear that I'm unable to solve are the following two :
1. We have an homotopy $H^n$ between $f_{|_{X^{n}}}$ and a map $X^n \longmapsto Y^n$.
I don't know how to construct the "global" homotopy gluing all the "local" homotopies, which should raise taking all the $f \circ \phi_{\alpha}$.
2.The homotopy $H^n$ is costant on $X^{n-1}$.
My thoughts:
In particular a specific homotopy $H_\alpha$ between $f \circ \phi_{\alpha}$ and $g_\alpha$ is a map from $E_\alpha^n \times I $ to $Y^n$ which is constant on $S_{\alpha}^{n-1}$. The problem here is that I can't compose with $\phi_{\alpha}^{-1}$ ($\phi_{\alpha}$ is an homeomorphism only of the interior part of $E_\alpha^n$, so the border of the disk would be "uncover") in order to obtain a map from a map from $X^n \times I$ to $Y^n$ which is what I want, since our final goal is to obtain a map homotopic to $f$ restricted to the $n$-skeleton.
I thought that maybe a map like $\begin{cases} H_\alpha \circ (\phi_{\alpha}^{-1}\times id) & x \in \phi_{\alpha}(\text{int}(E_{\alpha}^{n}) \\ f(x)\end{cases}$ could do the job (if continuos), but unfortunately this doesn't seem costant on $X^{n-1}$, required by $(3)$.
As far as concerns point $2$ I don't whether the property of being constant follows immediately from the costruction of $H^n$ in point $1$, but I suspect so. I probably have to use the fact that the "local" homotopies are costant on $S_{\alpha}^{n-1}$, so it requires $1.$
Notation edit: $f: X \longmapsto Y$ is a continuos map where $X,Y$ are $CW$ complex(with skeleton $X^n$ or $X_n$ depending on the book cited) with $(\phi_{\alpha})_{\alpha}$(or $\varphi_i$) the characteristic maps of the $CW$ complex. I'm denoting $E_{\alpha}^n$(or $\mathbb{D}^n$) and $S_{\alpha}^{n-1}$ as the $n-$euclidean closed disk and its boundary respectively.
$\rule{18cm}{1pt}$
Edit:
I think I found a proof of the fact that I'm looking for but I need to adjust the details. I've improved the previous homotopy given in Edit $1$ which was wrong.
The idea should follow Lemma 7.33 $p.187$ of Kirk and Davis lectures in algebraic topology, which can be found here. The proof is the following:

I don't get how $h_n$ is well defined. Of course is not defined on all $X_n$ in that way but only on $\varphi(\mathbb{D}^n)$ the image of the closed disk. So I thought that defining $h_n$ to be $\begin{cases}h_{n,i}(e) & e \in \mathbb{D}^n \\ h_{n-1}(x)\end{cases}$ could do the job but I'm not sure if this map is well defined and the continuity of this map, since the attaching maps need not to be injective on the boundary of $\mathbb{D}^n$.
Same problem costructing $h_{{n-1}_{|_{X_n}}}\simeq h_n$ where I tried (called $G$ the homotopy between $h_{n-1} \circ \varphi_i$ and $h_{n,i}$) $\begin{cases}G(\varphi_{i}^{-1}(x),t) & (x,t) \in \varphi_i(\text{int}(\mathbb{D}^{n})) \times I \\ h_{n-1}(x)\end{cases}$, which of course gives the desidered homotopy and is well defined (in the quotient space I don't think on the disjoint union) but I don't whether is continuos.
Which facts should I use in order to prove the continuity of such maps?
Any help,proof or tip would be appreciated, thanks in advance.
 A: This is a detail that's skimmed over in many proofs from what I can see. The idea is that:
(*) For a quotient map $p:A\rightarrow B$, a continuous map $H:A\rightarrow Y$ descends to a continuous map $H:B\rightarrow Y$ whenever $H$ is constant on all the fibers $p^{-1}(y)$.
In our case, let $J_n$ index the $n$-cells of $X$ and $D_j^n$ denote the $n$-disk of $j\in J_n$. We have a quotient map:
$$q_n:X^{n-1}\sqcup \bigsqcup_{j\in J_n} D_j^n\rightarrow X^n $$
To apply (*) to a homotopy, we need to see that:
$$q_n\times\text{id}_I:A=(X^{n-1}\times I)\sqcup (\bigsqcup_{j\in J_n} D_j^n\times I)\rightarrow B=X^n\times I $$
is a quotient map, where $I=[0,1]$. This can be shown to be the case for any compact space $I$.
For any $j\in J_n$, let $H_j:D_j^n\times I\rightarrow Y$ be a homotopy relative to $S_j^{n-1}$ between the map $h_{n-1}\circ\varphi_j$ and $h_{n,i}$. Define the map $H:A\rightarrow Y$ by:
$$H(x,t)=
\begin{cases}
H_j(x,t)\text{, if }x\in D_j^n\\
h_{n-1}(x)\text{, if }x\in X^{n-1}
\end{cases}
$$
Because $H_j$ is a homotopy relative to $S_j^{n-1}$, it keeps the values of $h_{n-1}\circ\varphi_j|_{S_j^{n-1}}$ fixed. Thus, by (*), $H$ descends to the desired continuous homotopy $X^n\times I \rightarrow Y$.
