$S^n$ is not a retract of the disk $D^{n+1}$ and Brouwer's Fixed Point Theorem. I was trying to understand Hirsch's proof of this fact:

"There is also a quick proof, by Morris Hirsch, based on the
  impossibility of a differentiable retraction. The indirect proof
  starts by noting that the map f can be approximated by a smooth map
  retaining the property of not fixing a point; this can be done by
  using the Weierstrass approximation theorem, for example. One then
  defines a retraction as above which must now be differentiable. Such a
  retraction must have a non-singular value, by Sard's theorem, which is
  also non-singular for the restriction to the boundary (which is just
  the identity). Thus the inverse image would be a 1-manifold with
  boundary. The boundary would have to contain at least two end points,
  both of which would have to lie on the boundary of the original
  ball—which is impossible in a retraction." - Wikipedia - Brouwer fixed-point theorem

But I am confused about usage of Sard's theorem. Why does it say about "1-manifold with boundary?" I thought Sard's gives us only "1-manifold." 
 A: Here is a slightly more general theorem on regular values.
Suppose $f: M \to N$ is a smooth function, where $M$ has boundary and $N$ does not. Suppose $p \in N$ is a regular value of $f$. Then $S = f^{-1}(p)$ is a "properly embedded" submanifold of $M$, which means precisely that $S \cap \partial M = \partial S$. This is proved in the exact same way as the theorem where $M$ has no boundary. (The reason $\partial S \subset \partial M$, for instance, is that if you apply the standard theorem to $f\big|_{\text{int } M}$, then you obtain a submanifold without boundary $S \cap \text{int }(M)$.)
Now because $D^{n+1}$ is compact and the above theorem, a regular value $p$ of a map $f: D^{n+1} \to S^n$ has $f^{-1}(p)$ a compact, properly embedded 1-manifold. Compact 1-manifolds have an even number of boundary points. But if $s \in f^{-1}(p)$ is a boundary point, because $f$ is a retraction, $f(s) = s$; so $s = p$. Because $p \in f^{-1}(p)$, we see that this compact 1-manifold would have a single boundary point. Nonsense.
