Proving the No Retraction Theorem I try to understand a proof of the No Retraction Theorem which states that for any compact smooth manifold $M$ with boundary $\partial M \neq \emptyset$, there is no smooth map 
$$
f : M \to \partial M , \ \ \ f|_{\partial M } = id.
$$
For the proof we suppose that such a map exists. Then by Sard's theorem we find a regular value $r \in \partial M$ and the preimage of $r$ under $f$ is a 1-dimensional manifold. The author now continues to say that the boundary of this preimage has to be a subset of $\partial M$. My questions is: Why do we know that?
 A: This is due to the form that the regular value theorem takes for manifolds with boundary:

Let $f:M\to N$ be a smooth map between manifolds with boundary and let $\partial N=\emptyset$. Let $c\in N$ be a regular value for $f$. Then, $f^{-1}(c)$ is a submanifold of $M$ such that $\dim f^{-1}(c)=\dim M-\dim N$ and $\partial f^{-1}(c)= f^{-1}(c)\cap \partial M$.

A: The conclusion of the Sard's Theorem argument is a bit stronger than what you say. 
The real conclusion is that the map $f : M \to \partial M$ has a regular value $r \in \partial M$. 
It follows, first, that $f^{-1}(r)$ is a 1-manifold with boundary. 
Now suppose that you take $x \in f^{-1}(r) \cap \text{interior}(M)$. The derivative map $D_x f : T_x M \to T_r M$ has rank $1$, so its kernel is a 1-dimensional subspace of $M$. Applying the implicit function theorem, and using that $x$ is in the interior of $M$, it follows that $f^{-1}(r)$ intersected with a small neighborhood of $x$ is a 1-manifold without boundary containing $r$. Thus, $x$ cannot be a boundary point of the 1-manifold $f^{-1}(r)$, therefore all of the boundary points of $f^{-1}(r)$ are contained in $\partial M$.
One can make a still stronger conclusion, namely that $f^{-1}(r)$ is a "properly embedded" submanifold of $M$, which means that each point on $\partial M \cap f^{-1}(r)$ has a neighborhood in which the intersection of that neighborhood with $f^{-1}(r)$ looks like the intersection of the half-disc $\{(x,y) \mid x^2 + y^2 < 0, |y| \ge 1\}$ with the line $x=0$.
