Normal Bundle of a Manifold I was reading "Morse Theory" by J.Milnor and at page number 32 there is remark
"It is not difficult that N is an n-dimensional manifold differentiably embedded in $\mathbb{R}^{2n}$ ( N is the total space of the normal vector bundle of M)"
If anyone can please justify claim .
 A: There are several equivalent ways to show a set is an embedded submanifold, and I think each allows a proof of that statement in a straightforward manner. One such approach would look as follows (with some of the details left to you):
(For the sake of completeness, here is the claim: $M\subset\mathbb{R}^n$ is assumed to be a $k$-dimensional  embedded submanifold, and 
$$N:=\{(q,v): q\in M, v  \text{ perpendicular to M at } q\}$$
Claim: $N$ is an embedded $n-$ dimensional submanifold of $\mathbb{R}^{2n}$).
By assumption, $M\subset \mathbb{R}^n$ is an embedded $k-$dimensional submanifold. This is equvialent to the statement that for $p\in M$ there is a neighbourhood $U$ of $p$ in $M\subset \mathbb{R}^n$ and a smooth map $f:U\rightarrow \mathbb{R}^{n-k}$ such that $rank(df)= n-k$ and $M\cap U = f^{-1}(0)$
Choose locally in $U$ $k$ smooth vectorfields $v_1,\ldots,v_k$ on $U$, which are tangent to $M$ when restricted to $M$ and span the tangent space of $M$ along $M$. The existence of such vectorfields is easy to prove (if you make $U$ smaller if necessary). 
Define a smooth map $F:U\times\mathbb{R}^n\rightarrow \mathbb{R}^n$ by
$$(p,v)\mapsto F(p,v):= (f(p), \langle v_1, v\rangle,\ldots, \langle v_k, v\rangle) $$
Then $N$ is the counterimage of $0$ under $F$, and the rank is easily seen to be $rank(df)+k=n$, since the $v_i$ are linearly independent, which then is true also for the linear functionals $v\mapsto \langle v_i,v \rangle$. Hence $N$ is an $n-$dimensional submanifold of $R^{2n}$
