I was reading Milnor's differential topology book,in the Brouwer degree chapter.there is a famous lemma state as follows:
If $f:M\to N$ extend to smooth map $F:X\to N$ then $\text{deg}(f,y) = 0$ for every regular value(where $\partial X =M$,all manifold are oriented,M is compact,N is connected)
- In the proof,we take the $F^{-1}(y)$ as regular value submanifold,it's one-dimension compact manifold,hence exist two cases arc and circle,in the book,case of arc is considered just prove:$\text{sgn}(df_a)+\text{sgn}(df_b) = 0$ ,I don't know why don't consider the case of circle.(since no intersection withh $\partial X$) is my understanding correct?
- the second question is the induced orientation for the arc,I don't understand why sgn of $df_{a/b}$ is induced by the map $F:X\to N$? I guess something to do with extension,but I can't make it clear
For $1$): Indeed the boundary points of $X$, $\partial X= M$ only intersect with the simple curves
For $2$): the orientation of an arc $A$ is induced by both $X$ and $N$: let $x\in A$ and let $(v_1,...,v_{n+1})$ be a positively oriented basis for the tangent space of $X$, with $v_1$ tangent to the simple curve $A$. Then the vector $v_1$ determines the appropriate orientation for the tangent space of the simple curve $A$ if the differential map $dF_x$ send $(v_2,...,v_{n+1})$ into a positively oriented basis for $N$. Hence the sign of $df_a$ induced by the map $F$.