Embedding a manifold in the disk I don't understand a sentence made by Hirsch in his Differential Topology at page 175:
If $k > n+1$ and $M^n = \partial W^{n+1}$, then an embedding $M^n \hookrightarrow S^{n+k}$ extends to a neat embedding $W^{n+1} \hookrightarrow D^{n+k+1}$.
($D^{n+k+1}$ is the disk of dimension $n+k+1$, $S^{n+k}$ is the boundary of $D^{n+k+1}$, i.e. the sphere of dimension $n+k$,  $W$ is a compact manifold of dimension $n+1$ and $M$ is the boundary of $W$.)
Could someone explain to me why this sentence is true?
Thanks to all!
 A: The proof of this has two steps.  
Step 1: Denote the map $M \to S^{n+k}$ by $f$.  Denote the inclusion $S^{n+k} \to D^{n+k+1}$ by $i$.   Then the map $i \circ f : M \to D^{n+k+1}$ extends to a smooth function $g : W \to D^{n+k+1}$.  You can define this extension in a variety of ways.  A natural extension would be to take a collar neighbourhood of $M$ in $W$, $\epsilon : M \times [0,1] \to W$, and then define $g$ inside the collar neighbourhood by $g(\epsilon(m,t)) = t^2m$ and outside the collar neighbourhood, define $g$ to be the zero vector.  Technically, you'll need to replace that $t^2$ by some $C^\infty$ increasing homeomorphism $[0,1]\to[0,1]$ such that its derivative at zero is zero.  With $t^2$ all you get is a $C^1$ embedding. If instead you use $t^3$ you have a $C^2$ embedding.  So I guess the function you need is $e^{1-\frac{1}{t^2}}$, with it defined to be zero at zero. I need my collar neighbourhood to be of the form $\epsilon(M \times \{1\}) = M = \partial W$.  
Step 2: Smooth approximation theory says you can approximate $g$ by an embedding (and since the map is already neat on the boundary, it will be a neat approximation).  The approximation theory only works if the ambient space has dimension strictly larger than twice the dimension of the domain, so you'll need $2(n+1)+1 \leq n+k+1$, equivalently $2n+3 \leq n+k+1$ or $n+2 \leq k$.  
And this is Hirsch's statement. 
