Problem Statement: Let $f:S^1\rightarrow S^1$ be a smooth map of manifolds where $S^1=\frac{[0,1]}{0~1}$, and let $f'(t)\in \mathbb{R}$ be given by the $df_t[1]_t=f'(t)[1]_{f(t)}$ at each $t\in S^1$. Let $v$ be a regular value of $f$
Rewrite the bounds of the integral using v, and conclude that $\int_0^1 f'(t)\ dt = n\in \mathbb{Z}$ and that $n$ is given by the signed sum of the pre-images of a regular value $v$.
I believe I have the proof that it's an integer. Under the mod $1$ relation $\int_0^1 f'(t)\ dt = 0$, hence the above integral (which is a real integral) must be equal to 0 mod 1, i.e, must be an integer. I didn't need any of the regular value business for this.
I'm having a little trouble on the next part. I've studied complex analysis and this is really similar to the argument principle, so I'm expecting that some how whenever I cross a value in the pre-image of $v$ while going from 0 to 1, I should get add one integer value to the integral. I know the number of pre-images of a regular value is the degree of the map $f$, and hence I intuitively expect that the result should be true, but I can't figure out how to prove that the integral increases by one each time it passes a regular value (or decreases, if the function isn't strictly monotonic).
A hint/method is preferred to the explicit answer. Thanks in advance, will upvote.
I don't think that the statement "the integral increases by one each time it passes a regular value" can be made both precise and true. Integrals of continuous functions don't jump like that.
Consider the universal cover $\pi:\mathbb R\to S^1$, $\pi(x)=x\mod 1$. The composition $f\circ \pi$ is $1$-periodic, and it lifts to a smooth map $F:\mathbb R\to\mathbb R$. Using this lift, you can argue along the following (mouse-over) lines:
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