This is a question about the proposition in p.71 of Guillemin&Pollack - Differential Topology.
Proposition. If $Y\subset \Bbb R^M$ then $N(Y)$ is a manifold of dimension $M$ and the projection $\sigma:N(Y)\to Y$ is a submersion.
Here $Y$ is an embedded submanifold in $\Bbb R^M$ and $N(Y)\subset Y\times \Bbb R^M$ is the total space of the normal bundle.
Proof. (given in the book) Define $Y$ locally by equations: around any given point of $Y$, find an open set $\tilde{U}$ of $\Bbb R^M$ and a submersion $\phi:\tilde{U}\to \Bbb R^k$ ($k=\text{codim}Y$) such that $U=Y\cap \tilde{U}=\phi^{-1}(0)$. The set $N(U)$ equals $N(Y)\cap (U\times \Bbb R^M)$, thus is open in $N(Y)$. For each $y\in U$, $d\phi_y:\Bbb R^M\to \Bbb R^k$ is surjective and has kernel $T_yY$. Therefore its transpose maps $\Bbb R^k$ isomorphically onto $N_yY$. The map $\psi:U\times \Bbb R^k\to N(U)$, defined by $\psi(y,v)=(y,(d\phi_y)^tv)$, is thus bijective, and it is easy to check that it is an embedding of $U\times \Bbb R^k$ into $Y\times \Bbb R^m$. ~~
How can we show that $\psi$ is an embedding? In this book an embedding means a map which is proper injective immersion. But it is quite clear that it suffices to show that the map $\psi$ is a diffeomorphism onto its image, which is a weaker condition than being an embedding. Also, it is clear from the text that $\psi$ is a smooth bijection (onto $N(U)$). However it doesn't seem easy to me to show that its inverse is also smooth. Thanks in advance.
Take a chart $y\colon U\to U'$ and promote it to charts $(y,\operatorname{id}_k)$ and $(y,\operatorname{id}_M)$ of $U\times\mathbb R^k$ and $U\times\mathbb R^M$. We get
where $$ \bar\psi(\mathbf y,\mathbf v)=(\mathbf y,L(\mathbf y)(\mathbf v)), $$ with $L(\mathbf y)$ a linear monomorphism for every $\mathbf y\in U'$. The Jacobian of $\bar\psi$ is therefore of the form $$ J(\bar\psi)=\begin{pmatrix} I &0\\ D &L(\mathbf y) \end{pmatrix} $$ for some matrix $D$ corresponding to partial derivatives of $L(\mathbf y)(\mathbf v)$ with respect to the coordinates $y_j$ of $\mathbf y$. Since $L(\mathbf y)$ is a monomorphism, $J(\bar\psi)$ is non-singular and so $\bar\psi$ and hence $\psi$ are immersions.