Normal bundle of a submanifold is again a manifold I'm studying differential topology by myself by reading Guillemin and Pollack's book and I encountered the following problem (Exercise 12 in Chapter 2.3):
Given $Z \subset Y \subset \mathbb{R}^M$, we consider the normal bundle to $Z$ in $Y$ defined as follows:
$$N(Z;Y) := \{(z,v): z \in Z, v \in T_{z}(Y) \text{ and } v \perp T_{z}(Z)\}$$
Show that the normal bundle $N(Z;Y)$ is a manifold with the same dimension as $Y$.
I have tried following the given hints to use Exercise 1.4.4 and proof of the Proposition on page 71 of the book. This gives us a parametrization $\phi: U \times \mathbb{R}^{l} \rightarrow N(Z;\mathbb{R}^{M}) = N(Z)$, which implies that the normal bundle $N(Z)$ is a manifold. However, I have been stuck on showing that this parametrization restricts to a parametrization $U \times \mathbb{R}^k \rightarrow N(Z;Y)$, which is suggested by the hint...
I have been thinking about this problem for almost a month without any significant progress, so any help/hints would be appreciated! Thank you so much in advance! Also, a post on the same question: Normal bundle to $Z$ in $Y$ is a manifold with same dimension as $Y$ (Exercise 2.3.12 of Guillemin-Pollack) (Though unfortunately there's no answer in this post...)
In addition, I have attached three images of the book below, just in case that they might be useful:
The problem along with the hint:

The claim proved in Exercise 1.4.4:

Proof of the Prop on page 71:

 A: When studying this theorem, I found it useful the concept of slice chart [cf. John M. Lee - Introduction to Smooth Manifolds]
Definition: A chart $\chi\colon U\to U'$ in an $n$-manifold $X$ is an $m$-slice chart for a subset $S\subseteq X$ if there exists a constant $\mathbf c\in\mathbb R^{n-m}$ and $n-m$ indexes $i_1,\dots,i_{n-m}$ such that
$$
   S\cap U = \{\pi\circ\chi=\mathbf c\},
$$
where $\pi$ is the projection $(x_1,\dots,x_n)\mapsto(x_{i_1},\dots,x_{i_{n-m}})$.
The relevant result here is the following
Lemma

*

*Every embedded $m$-submanifold $Y\subseteq X$ can be covered by $m$-slice charts for $Y$.


*Every topological subspace $Y\subseteq X$ that can be covered by $m$-slice charts for it has a smooth structure that makes it into an embedded submanifold of $X$.
Proof (sketch)

*

*Take charts $x\colon U\to U'$ and $y\colon Y\cap U\to V'$ such that, for $\bar\iota(x_1,\dots,x_m)=(x_1,\dots,x_m,0,\dots,0)$ we have




*Take an $m$-slice chart $x\colon U\to U'$ for $Y\cap U$. After reordering coordinates
$$
   Y\cap U = \{\pi\circ x=\mathbf c\},
$$
where $\pi(x_1,\dots,x_n)=(x_{m+1},\dots,x_n)$. Let $\bar\pi(x_1,\dots,x_n)=(x_1,\dots,x_m)$ be the complementary projection. Then, for $y=\bar\pi\circ x\circ\iota$ and $\jmath\colon(x_1,\dots,x_m)\mapsto(x_1,\dots,x_m,c_1,\dots,c_{n-m})$, we can use the following diagram to verify that (1) $y$ is a bijection and (2) any transition between two of these bijections is smooth.



Theorem. If $X$ is an $m$-submanifold embedded in $\mathbb R^n$, its normal bundle $\mathbf N(X)$ has a structure of $n$-submanifold embedded in $\mathbb R^{2n}$.
Proof (sketch)
By definition $\mathbf N(X)$ is a topological subspace of $X\times\mathbb R^n$. Therefore, it is enough to show that $\mathbf N(X)$ can be covered with $m$-slice charts. The idea is to take a slice chart $\chi\colon U\to U'$ for $X$ and produce a slice chart for $\mathbf N(X)$.
We may assume that $U\cap X=\{\pi\circ\chi=\mathbf0\}$. Put $\varphi=\pi\circ\chi$. Then $d\varphi(p)\colon\mathbb R^n\to\mathbb R^{n-m}$ satisfies
$$
   \textrm{im}(d\varphi(p)^T)=T_p(X)^\perp\tag1
$$
To see this observe that $T_p(X)=\ker(d\varphi(p))$ because $\subseteq$ + $=$ dim. Then $d\varphi(p)^T$ is mono with image $T_p(X)^\perp$, again $\subseteq$ + $=$ dim.
Next observe that
$$
   d\varphi(p)^T = d\chi(p)^T\circ\iota,\tag2
$$
where $\iota(z_{m+1},\dots,z_n)=(0,\dots,0,z_{m+1},\dots, z_n)$ is the transpose of $\pi$.
We can now prove that
\begin{align*}
    \Phi\colon U\times\mathbb R^n&\to U'\times\mathbb R^n\\
    (p,\mathbf w)&\mapsto(\chi(p),(d\chi(p)^T)^{-1}(\mathbf w)).
\end{align*}
defines an $n$-slice chart for $\mathbf N(U\cap X)=\mathbf N(X)\cap U\times\mathbb R^n$.
To see this observe, using $(1)$, that
$$
    (p,\mathbf w)\in \mathbf N(U\cap X)\iff \chi(p)\in\{\bar\pi=\mathbf0\}
        \textrm{ and }\mathbf w\in\textrm{im}(d\varphi(p)^T),\tag3
$$
where $\bar\pi\colon U'\to\pi(U')\subseteq\pi\colon(x_1,\dots,x_n)\mapsto(x_{m+1},\dots,x_n)$.
Therefore, for $(q,\mathbf v)=\Phi(p,\mathbf w)$, we have $q=\chi(p)$ and $d\chi(p)^T(\mathbf v)=\mathbf w$. Then
\begin{align*}
    (q,\mathbf v)\in\Phi(\mathbf N(U\cap X)) &\iff (p,\mathbf w)\in \mathbf N(U\cap X)\\
        &\iff \chi(p)\in\{\bar\pi=\mathbf0\}
            \textrm{ and }\mathbf w\in\textrm{im}(d\chi(p)^T\circ\iota)
               &&\textrm{; by }(2)\textrm{ and }(3)\\
        &\iff q\in\{\pi=\mathbf0\}\cap U'\textrm{ and }\mathbf v\in\textrm{im}(\iota),
\end{align*}
which completes the proof because
$$
    2n-\dim\{\pi=0\}-\dim\textrm{im}(\iota)= 2n - m - (n-m) = n.
$$
