Proving the fundamental group of circle is infinite.

Question

I am trying the prove that fundamental group of circle is infinite by using the concept of degree of loops.

Here I am giving the details of what definitions I am using here.

$$S^{1}:=\{ z \in \mathbb{C} | |Z|=1 \}$$

Let $f$ be loop in $S^{1}$ based at 1 and $P=\{0=x_0,x_1,x_2,...,x_n=1 \}$ be a partition of $[0,1]$ such that $x,x^{'} \in [x_{i-1},x_i] \implies |f(x)-f(x^{'})|<1$

$\alpha(\frac{f(x_i)}{f(x_{i-1})} )$ be the unique argument of the complex number $\frac{f(x_i)}{f(x_{i-1})}$ in the interval $(-\frac{\pi}{2},\frac{\pi}{2})$

$$deg(f):=\frac{1}{2\pi}\sum_{i=1}^{n}\alpha(\frac{f(x_i)}{f(x_{i-1 })})$$

Below are some things I Have proved

1. Degree is independent of partition

2. such an $\alpha$ exists

3. degree is an integer

Now I want to prove that if two loops in $S^{1}$ based at 1 are path homotopic then $deg(f)=deg(g)$

Let $F$ be a path homotopy between $f$ and $g.$ i.e. $F:[0,1] \times [0,1] \rightarrow S^{1}$ such that $F(x,0)=f(x); F(x,1)=g(x) \forall x \in [0,1]$ and $F(0,t)=F(1,t)=1 \forall x \in [0,1].$

Since $F$ is continuous, and $[0,1] \times [0,1]$ is compact space, $F$ is uniformly continuous, So i can choose partitions $\{0=x_0,x_1,...,x_n=1\}$ and $\{0=t_1,t_2,...t_m=1 \}$ such that $(x,t),(x^{'},t{'}) \in [x_{i-1},x_i] \times [t_{j-1},t_j] \implies |F(x,t)-F(x^{'},t^{'})|<1$ From above it can be concluded that $x,x^{'} \in [x_{i-1},x_i] \implies |f(x)-f(x^{'})|<1 $ and $t,t^{'} \in [0,1] \implies |g(t)-g(t^{'})|<1$

$\implies deg(f)=\frac{1}{2\pi}\sum_{i=1}^{n}\alpha(\frac{f(x_i)}{f(x_{i-1})})$ and $deg(g)=\frac{1}{2\pi}\sum_{j=1}^{m}\alpha(\frac{g(t_i)}{g(t_{i-1})})$

Now it is enough to show that $\sum_{i=1}^{n}\alpha(\frac{f(x_i)}{f(x_{i-1})})=\sum_{j=1}^{m}\alpha(\frac{g(t_i)}{g(t_{i-1})})$

From here I don't know where to go.

Answer 1

You have proved that the number $\deg(f)$ is well-defined.

Let us analyze the function $\alpha$ occurring in the degree-formula. For each $w \in S^1_+ = \{ z \in S^1 \mid \Re (z) > 0\}$ we can define $\alpha(w) \in (-\frac \pi 2, \frac \pi 2)$ to be the unique number such that $e^{i\alpha(w)} = w$.

1. For $w \in S^1$ we have $w\overline w = 1$, thus $\frac 1 w = \overline w$. Hence for $w \in S^1_+$ we get $\frac 1 w \in S^1_+$ and $$\alpha(\frac 1 w) = -\alpha(w) .\tag{1}$$

2. If $w_1, w_2 \in S^1_+$ and $w_1w_2 \in S^1_+$, then $$\alpha(w_1w_2) = \alpha(w_1) + \alpha(w_2). \tag{2}$$ In fact, we have $w_k = e^{i\alpha(w_k)}$ and $w_1 w_2 = e^{i\alpha(w_1 w_2)}$, hence $\alpha(w_1) + \alpha(w_2) = \alpha(w_1w_2) + 2r\pi$ for some $r \in \mathbb Z$. But $\alpha(w_1) + \alpha(w_2) \in (-\pi, \pi)$ and $\alpha(w_1w_2) +2r\pi \in J_r := (2r\pi - \frac \pi 2, 2r\pi + \frac \pi 2)$. Since $(-\pi, \pi) \cap J_r = \emptyset$ for $r \ne 0$, we must have $r = 0$ which proves (2).

3. If $z, z' \in S^1$, then $\lvert z - z' \rvert < 1$ implies $\frac{z}{z'} \in S^1_+$. Thus $\alpha(\frac{z}{z'})$ is well-defined.

Given a path homotopy $F$ from $f$ to $g$, you found partitions $\{0=x_0,x_1,...,x_n=1\}$ and $\{0=t_1,t_2,...t_m=1 \}$ such that $$(x,t),(x',t') \in [x_{i-1},x_i] \times [t_{j-1},t_j] \implies |F(x,t)-F(x',t')|<1 . \tag{3}$$

Define $f_j : S^1 \to S^1, f_j(x) = F(x,t_j)$. These are loops based at $1$. Since $f_1 = f$ and $f_m = g$, it suffices to show that $\deg(f_{j-1}) = \deg(f_j)$.

From (3) we know that $x, x' \in [x_{i-1},x_i] \implies |f_j(x) - f_j(x')| < 1$ for all $j$. Thus for all $j$ $$\deg(f_j) = \frac{1}{2\pi}\sum_{i=1}^n\alpha(\frac{f_j(x_i)}{f_j(x_{i-1})}) .$$

We claim that $$\alpha(\frac{f_{j-1}(x_i)}{f_{j-1}(x_{i-1})}) = \alpha(\frac{f_j(x_{i-1})}{f_{j-1}(x_{i-1})}) + \alpha(\frac{f_j(x_i)}{f_j(x_{i-1})}) + \alpha(\frac{f_{j-1}(x_i)}{f_j(x_i)}) \tag{4}.$$ Let us defer the proof of (4). Using (1) we get $$\alpha(\frac{f_{j-1}(x_i)}{f_{j-1}(x_{i-1})}) = \alpha(\frac{f_j(x_{i-1})}{f_{j-1}(x_{i-1})}) + \alpha(\frac{f_j(x_i)}{f_j(x_{i-1})}) - \alpha(\frac{f_j(x_i)}{f_{j-1}(x_i)}) \tag{5}.$$

We conclude

$$2\pi\deg(f_{j-1}) = \sum_{i=1}^n \alpha(\frac{f_{j-1}(x_i)}{f_{j-1}(x_{i-1})}) = \sum_{i=1}^n ( \alpha(\frac{f_j(x_{i-1})}{f_{j-1}(x_{i-1})}) + \alpha(\frac{f_j(x_i)}{f_j(x_{i-1})}) - \alpha(\frac{f_j(x_i)}{f_{j-1}(x_i)}) ) \\ = \sum_{i=1}^n \alpha(\frac{f_j(x_{i-1})}{f_{j-1}(x_{i-1})}) + \sum_{i=1}^n \alpha(\frac{f_j(x_i)}{f_j(x_{i-1})}) - \sum_{i=1}^n \alpha(\frac{f_j(x_i)}{f_{j-1}(x_i)})\\ = 2\pi \deg(f_j) + \sum_{i=1}^n \alpha(\frac{f_j(x_{i-1})}{f_{j-1}(x_{i-1})}) - \sum_{i=1}^n \alpha(\frac{f_j(x_i)}{f_{j-1}(x_i)}) = 2\pi \deg(f_j) + \alpha(\frac{f_j(x_0)}{f_{j-1}(x_0)}) - \alpha(\frac{f_j(x_n)}{f_{j-1}(x_n)}) \\ = 2\pi \deg(f_j)$$ because $f_k(x_0) = f_k(0) = 1 = f_k(1) = f_k(x_n)$.

We come to the proof of (3).

Let $z_1 = f_{j-1}(x_{i-1}) = F(x_{i-1},t_{j-1})$, $z_2 = f_j(x_{i-1}) =F(x_{i-1},t_j)$, $z_3 = f_j(x_i) = F(x_i,t_j)$, $z_4 = f_{j-1}(x_i) = F(x_i,t_{j-1})$. By (3) we have $\lvert z_j - z_k \rvert < 1$, thus all quotients $$\frac{z_j}{z_k} \in S^1_+ .$$ By (2) we get $$\alpha(\frac{z_4}{z_1}) = \alpha(\frac{z_4}{z_3}\frac{z_3}{z_1}) = \alpha(\frac{z_4}{z_3}) + \alpha(\frac{z_3}{z_1}) = \alpha(\frac{z_4}{z_3}) + \alpha(\frac{z_3}{z_2}\frac{z_2}{z_1}) = \alpha(\frac{z_4}{z_3}) + \alpha(\frac{z_3}{z_2}) + \alpha(\frac{z_2}{z_1}) .$$ This is nothing else than (4).