I'm trying to solve the following problem
Let $f: S^1 \to \mathbb{R}$ be a smooth Morse function, then $f$ has an even number of critical point.
My progress:
I was able to prove that if $g:\mathbb{R}\to \mathbb{R}$ is a Morse function with finite critical points $t_1 < ...< t_n$ then the index of the critical points must alternate (the details of this affirmation can be seen in the answer of this question), i.e; $$\text{sgn} (g'' (t_i)) =(-1)^{n-1} \text{sgn} (g'' (t_1)). $$
If $ x_1,..., x_k$ are the critical points of $f$, then we can define the diffeomorphism
$$\varphi: S^1\setminus\{x_1\} \to \mathbb{R}, $$
where $\varphi$ is a diffeomorphism that "looks like" the stereographic projection.
Then $g := f\circ\varphi^{-1} : \mathbb{R}\to \mathbb{R}$, is a Morse function. If we suppose by reductio ad absurdum, that $g$ has an even number of critical points $s_1<...<s_{2k}$, then we would conclude that
$$ \text{sgn}(g'(-\infty,s_1)) = (-1)^{2k}\text{sgn}(g'(s_{2k},\infty)) = \text{sgn}(g'(s_{2k},\infty)) $$
I think that this implies that $x_1$ is a degenerated point of $f$, but I wasn't able to show this.
Can anyone help me?
If $ S^1 =\{ e^{it}|0\leq t <2\pi\}$, then assume that $f'(0)=0$. So $$ 0=t_0<t_1<t_2<\cdots < t_n=2\pi $$
where $f'(t_n)=0$. If $f$ is increasing on $[0,t_1]$, then $f$ is decreasing on $[t_1,t_2]$. That is, $f$ is decreasing on $[t_{n-1},t_n]$ so that increasing and decreasing intervals have same number.