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This was a proposed exercise in some lecture notes on differential topology I’m reading. Some proofs for this result are given in the answers to this question. However, they assume previous knowledge of algebraic topology, which the lecture notes don’t. That’s why I was curious to know if there’s an alternative proof.

The definition of degree of a map I’m using is the following: Let $f:X \rightarrow Y$ be a smooth map between two oriented manifolds of the same dimension. Let $y\in Y$. Then $\deg_y(f)=I(f,\{y\})$, where $I(f,\{y\})$ is the intersection number of $f$ with $\{y\}$. If we also have that $f^{-1}(y)$ is a finite set $\forall y \in Y$ and that $Y$ is a connected manifold, then we define $\deg(f)=\deg_y(f)$ for any $y \in Y$.

I know as well that if $f,g:X \rightarrow Y$ are smooth maps such that $X$ is a compact manifold, $Y$ is a connected manifold and $f$ is homotopic to $g$, then $\deg(f)=\deg(g)$. I was thinking we could maybe use this in addition to the fact that the map $z\rightarrow z^n$ is homotopic to the identity map concatenated with itself $n$ times to prove that $\deg(f)=n$ (where $f: S^1 \to S^1$ is such that $f(z) = z^n$)?

dahemar
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    Not only is the map $z \mapsto z^n$ homotopic to the identity map concatenated with itself $n$ times, but those two maps are the same map. – Lee Mosher May 12 '21 at 12:58
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    It seems to me that you can apply your definition directly: starting with the map $f(z)=z^n$; use $y=1 + 0i \in S^2$; compute $f^{-1}(y)$ to be $n$ equally spaced points around $S^1$ (the $n^{\text{th}}$ roots of unity); and compute each of those individual points to have intersection number $1$, hence $I(f,{y}) = n \cdot 1 = n$. – Lee Mosher May 12 '21 at 13:09
  • @LeeMosher Thank you! Then, I must see that for every root of unity $x \in f^{-1}(y)$, the orientation of $df_x (T_x S^1) \bigoplus T_{f(x)}{y}$ matches the orientation of $T_{f(x)}{S^1}$? – dahemar May 12 '21 at 19:48
  • What you wrote looks a little garbled: You must show that the $df_x$-image of the given orientation on $T_x S^1$, which is an orientation on $df_x(T_x S^1) = T_{f(x)}(S^1) = T_y(S^1)$, matches the given orientation on $T_y S^1$. – Lee Mosher May 12 '21 at 20:20
  • @LeeMosher Thank you! I was just trying to use the definition of intersection number as a sum of orientation numbers that was given in the lecture notes (which also appears here for instance, page 7). But it must be equivalent to the one you mentioned. – dahemar May 12 '21 at 20:38
  • is there perhaps a way to use riemann surfaces here? @LeeMosher afaik $S^2$ is riemann surface and then...hmmm...this is $S^1$ though. but anyway great question. homology. come on. kinda forgot about degree for smooth (i last learned degree for riemann surfaces), but i kinda remember homology doesn't sound natural to answer this – BCLC May 13 '21 at 03:41

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I'll turn my comments into an answer.

You can apply the definition of degree directly. Starting with the map $f(z)=z^n$, and using $y = 1 + 0i \in S^1$, the set $f^{-1}(y)$ consists of $n$ equally spaced points around $S^1$, namely the $n^{\text{th}}$ roots of unity, $x = \exp(2 \pi i k/n)$, $k=0,...,n-1$.

So, for each $x \in f^{-1}(y)$, you just have to prove that the local degree of $f$ at $x$ is equal to $+1$, because then the global degree is $I(f,\{y\}) = n \cdot 1 = n$. In fact it's just as easy to ignore the special value $y=1+0i$, and prove this local degree fact for all $x \in S^1$ and $y=f(x) \in S^1$.

What this local degree computation comes down to is to prove that the map $$df_x : T_x S^1 \to T_{f(x)}S^1 = T_y S^1 $$ takes the standard orientation on $T_x S^1$ to the standard orientation on $T_y S^1$.

Lee Mosher
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