Linking number of two manifolds The following is part of Problem 12 in Milnor's Topology from the Differentiable Viewpoint. Some context is needed from Problem 11, which I will give here. Let $M$ be a manifold embedded in $\mathbb{R}^k$.
From Problem 11: The normal bundle space $$E=\{(x,v)\in M\times \mathbb{R}^k\;|\;v\perp TM_x\}$$ is a smooth manifold. If $M$ is compact and boundaryless, then the correspondence $$(x,v)\mapsto x+v$$ from $E$ to $\mathbb{R}^k$ maps some $\epsilon$-neighborhood of $M\times 0$ in $E$ diffeomorphically onto the $\epsilon$-neighborhood $N_\epsilon$ of $M$ in $\mathbb{R}^k$.
Problem 12: Define $r:N_\epsilon\to M$ by $r(x+v)=x$. Show that $r(x+v)$ is closer to $x+v$ than any other point of $M$.
I have done the later part of the Problem 12, my question is how to do the part I have stated here. Intuitively, this makes perfect sense, but I am not sure hot to justify the statement by using what is either given or presupposed in the book. Any help would be appreciated.
 A: Let me name some of the things in your post:


*

*$E_\epsilon$ is the $\epsilon$ neighborhood of $M \times 0$ in $E$

*$f : E_\epsilon \to N_\epsilon$ is the diffeomorphism given by the formula $f(x,v) = x+v$.


Suppose that there exists a point in $M$ which is closer to $x+v$ than $x=r(x+v)$ itself. By compactness of $M$, there exists a point $p \in M$ which minimizes the distance to $x+v$, and it follows that 
$$d(x+v,p) < d(x+v,r(x+v)) = |v|
$$
Consider the segment $\overline{x+v,p}$. Because $p \in M$ minimizes the distance to $x+v$, this segment meets $M$ orthogonally at the point $p$. Geometrically this is kind of obvious. For a rigorous proof using calculus, just apply the Lagrange multiplier method of constrained optimization.
It follows that 
$$f(x,v) = x+v = p+w = f(p,w)
$$
for some vector $w$ such that $w \perp T_p M$ and 
$$|w| = d(p+w,p) = d(x+v,p) < |v| < \epsilon
$$
Thus both of $(x,v)$ and $(p,w)$ are contained in $E_\epsilon$, hence $f$ is not one-to-one, a contradiction. 
