Why is $f(U) \cap V$ the zero set of $y^{n+1}, \dots, y^m$? I'm studying Tu's proof (p. 123) of theorem 11.13, but I just have a question about one detail.
He has that $f\colon N\to M$ is an embedding of a manifold of dimension $n$ in a manifold of dimension $m$, and he shows that given $p\in N$, there are local coordinates $(U, x^1, \dots, x^n)$ near $p$ and $(V, y^1, \dots, y^m)$ near $f(p)$ such that $f\colon U\to V$ has the form
$$
(x^1, \dots, x^n) \mapsto (x^1, \dots, x^n, 0, \dots, 0).
$$
He then says

Thus, $f(U)$ is defined in $V$ by the vanishing of the coordinates $y^{n+1}, \dots, y^m$.

I understand this statement to mean that
$$
f(U) \cap V = \{ q\in V \mid y^{n+1}(q) = \dots = y^m(q) = 0\}.
$$
However, I don't understand why this should be the case. Couldn't there be other points of $V$ at which the coordinates $y^{n+1}, \dots, y^m$ vanish?
 A: This is a consequence of the constant rank theorem.
https://fr.wikipedia.org/wiki/Th%C3%A9or%C3%A8me_du_rang_constant
I outline a proof here:
Since $N$ is a submanifold of $M$, for every $x\in N$, there exists a neighborhood $U$ of $x$ in $M$ (that you identify with a subset of $R^n$) and a submersion $g:U\rightarrow R^{n-p}$ such that $g^{-1}(0)=N\cap U$. You may assume that the restriction of $dg_0$ to $Vect(e_1,...,e_{n-p})$ is an isomorphism. Where $(e_1,...,e_n)$ is a base of $R^n$. Let $h: U\rightarrow R^n$ defined by $h(x_1,...,x_n)= (x_{n-p+1},..x_n,g(x_1,..,x_n))$   $dh_{0,..0,x}$ is an isomorphism. You can apply the inverse function theorem and suppose that $h:U\rightarrow V$ is invertible by eventually shrinking $U$. Then define $f=h^{-1}$ are your local coordinates.
You have $f(x_1,..,x_p,0...0)= N\cap U$.
A: You know that given $f:N\to M$, there is a chart $(x,U)$ of $p$ and $(y,V)$ of $f(p)$ such that $y\circ f\circ x^{-1}:\Bbb R^n\to \Bbb R^m$ has the form $(a^1,\ldots,a^n)\to (a^1,\ldots,a^n,0,\ldots,0)$. 
Now pick $q$ in $V\subseteq M$; and suppose that $y(q)$ has $0$ "last" coordinates, that is $y(q) = (a^1,\ldots,a^n,0,\ldots,0)$. By the above, $yfx^{-1}(a^1,\ldots,a^n) = y(q)$, so $q=fx^{-1}(a^1,\ldots,a^n)\in f(U)$.
Conversely if we pick $q\in f(U)\cap V$ the above gives $q$ has zero last $y$-coordinates. 
