Brouwer degree extension Lemma

Question

Let $M$ and $N$ be oriented $n$-dimensional manifolds without boundary an also $M$ is compact and $N$ connected. Suppose that $M$ is the boundary of a compact oriented manifold $X$ and that $M$ is oriented as the boundary of $X$. If $f: M \to N$ extends to a smooth map $F: X \to N$, then $\deg (f;y)=0$ for every regular value $y$.

In the proof of Lemma it supposed that $y$ is a regular value for $F$, so $F^{-1}(y)$ is compact $1$-dimensional manifold. And now I don't understand the following assertion:

"$F^{-1}(y)$ is a finite union of arcs and circles, with only the boundary points of the arcs lying on $M= \partial X$".

I guess that we cocnclude the finite union by the compactness of $F^{-1}(y)$ but I'm not sure how to prove it. And also I don't know how to prove that it's impossible to exist an arc with no boundary points on $M$. Thank you very much.

Answer 1

If $y$ is a regular value for both $f$ and $F$ then $F^{-1}(y)$ is a one-dimensional manifold whose boundary is $F^{-1}(y) \cap \partial X = F^{-1}(y) \cap M = f^{-1}(y)$. Since $X$ is compact and $F^{-1}(y)$ is closed, it follows that $F^{-1}(y)$ is compact. The classification of compact one dimensional manifolds with boundary implies that they are a finite union of closed intervals (for which the boundary consists of two points) and circles (that have no boundary) and since the boundary of $F^{-1}(y)$ lies on $M$, the closed intervals (arcs) of $F^{-1}(y)$, if there are any, must start and end in $M$.

Answer 2

The following generalization of the regular values theorem is standard:

Let $F:X \rightarrow N$ with $X$ a manifold with non-empty boundary and $N$ a manifold with boundary (maybe empty). If $y \in N - \partial N$ is a regular value for both $F$ and $F|_{\partial X}$ then $F^{-1}(y)$ is a neat submanifold of $X$, where a neat submanifold $A$ of $X$ is a submanifold such that $\partial A = \partial X \cap A$. You can find this theorem for instance in Hirsch's "Differential topology", Chapter 1, Theorem 4.2.

Therefore, in your situation, where the hypothesis of the lemma are satisfied, you have that $F^{-1}(y)$ is a neat 1-dimensional submanifold with boundary of $X$. In particular, the boundary of $F^{-1}(y)$ lies in $\partial X = M$. The rest follows as you said from compactness and the fact that the only compact 1-dimensional manifolds with boundary are the circle and the closed interval.