4.3 Examples of fundamental groups

There does not exist a routine procedure to compute the fundamental groups, in general. However, in certain cases, they are obtained by relatively simple considerations. Here we look at the fundamental groups of the circle $S^1$ and related spaces.

Let us express $S^1$ as $\{z\in \mathbb{C}||z|=1\}$ . Define a map $p:\mathbb{R}\to S^1$ by $p:x\mapsto \exp (\mathrm{i}x)$ . Under $p$ , the point $0\in \mathbb{R}$ is mapped to $1\in S^1$ , which is taken to be the base point. We imagine that $\mathbb{R}$ wraps around $S^1$ under $p$ , see figure 4.9. If $x,y\in \mathbb{R}$ satisfies $x-y=2\pi m(m\in \mathbb{Z})$ , they are mapped to the same point in $S^1$ . Then we write $x\sim y$ . This is an equivalence relation and the equivalence class $[x]=\{y|x-y=2\pi m\text{ for some }m\in \mathbb{Z}\}$ is identified with a point $\exp (\mathrm{i}x)\in S^1$ . It then follows that $S^1\cong \mathbb{R}/2\pi \mathbb{Z}$ . Let $\tilde{f}:\mathbb{R}\to \mathbb{R}$ be a continuous map such that $\tilde{f}(0)=0$ and $\tilde{f}(x+2\pi )\sim \tilde{f}(x)$ . It is obvious that $\tilde{f}(x+2\pi )=\tilde{f}(x)+2n\pi$ for any $x\in \mathbb{R}$ , where $n$ is a fixed integer. If $x\sim y(x-y=2\pi m)$ , we have

Figure 4.9. The map $p:\mathbb{R}\to S^1$ defined by $x\mapsto \exp (\mathrm{i}x)$ projects $x+2m\pi$ to the same point on $S^1$ , while $\tilde{f}:\mathbb{R}\to \mathbb{R}$ , such that $\tilde{f}(0)=0$ and $\tilde{f}(x+2\pi )=\tilde{f}(x)+2n\pi$ for fixed $n$ , defines a map $f:S^1\to S^1$ . The integer $n$ specifies the homotopy class to which $f$ belongs.

$$\begin{array}{rl}\tilde{f}(x)-\tilde{f}(y)& =\tilde{f}(y+2\pi m)-\tilde{f}(y)\\ & =\tilde{f}(y)+2\pi mn-\tilde{f}(y)=2\pi mn\end{array}$$

hence $\tilde{f}(x)\sim \tilde{f}(y)$ . Accordingly, $\tilde{f}:\mathbb{R}\to \mathbb{R}$ uniquely defines a continuous map $f:\mathbb{R}/2\pi \mathbb{Z}\to \mathbb{R}/2\pi \mathbb{Z}$ by $f([x])=p\circ \tilde{f}(x)$ , see figure 4.9. Note that $f$ keeps the base point $1\in S^1$ fixed. Conversely, given a map $f:S^1\to S^1$ , which leaves $1\in S^1$ fixed, we may define a map $\tilde{f}:\mathbb{R}\to \mathbb{R}$ such that $\tilde{f}(0)=0$ and $\tilde{f}(x+2\pi )=\tilde{f}(x)+2\pi n$ .

In summary, there is a one-to-one correspondence between the set of maps from $S^1$ to $S^1$ with $f(1)=1$ and the set of maps from $\mathbb{R}$ to $\mathbb{R}$ such that $\tilde{f}(0)=0$ and $\tilde{f}(x+2\pi )=\tilde{f}(x)+2\pi n$ . The integer $n$ is called the degree of $f$ and is denoted by $\mathrm{deg}(f)$ . While $x$ encircles $S^1$ once, $f(x)$ encircles $S^1$ times.

Lemma 4.2

(1) Let $f,g:S^1\to S^1$ such that $f(1)=g(1)=1$ . Then $\mathrm{deg}(f)=\mathrm{deg}(g)$ if and only if $f$ is homotopic to $g$ .

(2) For any $n\in \mathbb{Z}$ , there exists a map $f:S^1\to S^1$ such that $\mathrm{deg}(f)=n$ .

Proof

(1) Let $\mathrm{deg}(f)=\mathrm{deg}(g)$ and $\tilde{f},\tilde{g}:\mathbb{R}\to \mathbb{R}$ be the corresponding maps. Then $\tilde{F}(x,t)\equiv t\tilde{f}(x)+(1-t)\tilde{g}(x)$ is a homotopy between $\tilde{f}(x)$ and $\tilde{g}(x)$ . It is easy to verify that $F\equiv p\circ \tilde{F}$ is a homotopy between $f$ and $g$ . Conversely, if $f\sim g:S^1\to S^1$ , there exists a homotopy $F:S^1\times I\to S^1$ such that $F(1,t)=1$ for any $t\in I$ . The corresponding homotopy $\tilde{F}:\mathbb{R}\times I\to \mathbb{R}$ between $\tilde{f}$ and $\tilde{g}$ satisfies $\tilde{F}(x+2\pi ,t)=\tilde{F}(x,t)+2n\pi$ for some $n\in \mathbb{Z}$ . Thus, $\mathrm{deg}(f)=\mathrm{deg}(g)$ .

(2) $\tilde{f}:x\mapsto nx$ induces a map $f:S^1\to S^1$ with $\mathrm{deg}(f)=n$

Lemma 4.2 tells us that by assigning an integer $\mathrm{deg}(f)$ to a map $f:S^1\to S^1$ such that $f(1)=1$ , there is a bijection between ${\pi }_1(S^1,1)$ and $\mathbb{Z}$ . Moreover, this is an isomorphism. In fact, for $f,g:S^1\to S^1$ , $f\ast g$ , defined as a product of loops, satisfies $\mathrm{deg}(f\ast g)=\mathrm{deg}(f)+\mathrm{deg}(g)$ . [Let $\tilde{f}(x+2\pi )=\tilde{f}(x)+2\pi n$ and $\tilde{g}(x+2\pi )=\tilde{g}(x)+2\pi m$ . Then $f\ast g(x+2\pi )=f\ast g(x)+2\pi (m+n)$ . Note that $\ast$ is not a composite of maps but a product of paths.] We have finally proved the following theorem.

Theorem 4.5

The fundamental group of $S^1$ is isomorphic to $\mathbb{Z}$ ,

$${\pi }_1\left(S^1\right)\cong \mathbb{Z}.\text{\hspace{1em}(4.10)}$$

[Since $S^1$ is arcwise connected, we may drop the base point.]

Although the proof of the theorem is not too obvious, the statement itself is easily understood even by children. Suppose we encircle a cylinder with an elastic band. If it encircles the cylinder $n$ times, the configuration cannot be continuously deformed into that with $m(\ne n)$ encirclements. If an elastic band encircles a cylinder first $n$ times and then $m$ times, it encircles the cylinder $n+m$ times in total.

4.3.1 Fundamental group of torus

Theorem 4.6

Let $X$ and $Y$ be arcwise connected topological spaces. Then ${\pi }_1(X\times Y,(x_0,y_0))$ is isomorphic to ${\pi }_1(X,x_0)\oplus {\pi }_1(Y,y_0)$ .

Proof

Define projections $p_1:X\times Y\to X$ and $p_2:X\times Y\to Y$ . If $\alpha$ is a loop in $X\times Y$ at $(x_0,y_0)$ , ${\alpha }_1\equiv p_1(\alpha )$ is a loop in $X$ at $x_0$ , and ${\alpha }_2\equiv p_2(\alpha )$ is a loop in $Y$ at $y_0$ . Conversely, any pair of loops ${\alpha }_1$ of $X$ at $x_0$ and ${\alpha }_2$ of $Y$ at $y_0$ determines a unique loop $\alpha =({\alpha }_1,{\alpha }_2)$ of $X\times Y$ at $(x_0,y_0)$ . Define a homomorphism $\phi :{\pi }_1(X\times Y,(x_0,y_0))\to {\pi }_1(X,x_0)\oplus {\pi }_1(Y,y_0)$ by

$$\phi ([\alpha ])=([{\alpha }_1],[{\alpha }_2]).$$

By construction $\phi$ has an inverse, hence it is the required isomorphism and ${\pi }_1(X\times Y,(x_0,y_0))\cong {\pi }_1(X,x_0)\oplus {\pi }_1(Y,y_0)$ .

Example 4.3

(1) Let $T^2=S^1\times S^1$ be a torus. Then

$${\pi }_1\left(T^2\right)\cong {\pi }_1\left(S^1\right)\oplus {\pi }_1\left(S^1\right)\cong \mathbb{Z}\oplus \mathbb{Z}.\text{\hspace{1em}(4.11)}$$

Similarly, for the $n$ -dimensional torus

$$T^n=\underset{n}{\underbrace{S^1\times S^1\times \cdots \times S^1}}$$

we have

$${\pi }_1\left(T^n\right)\cong \underset{n}{\underbrace{\mathbb{Z}\oplus \mathbb{Z}\oplus \cdots \oplus \mathbb{Z}}}.\text{\hspace{1em}(4.12)}$$

(2) Let $X=S^1\times \mathbb{R}$ be a cylinder. Since ${\pi }_1(\mathbb{R})\cong \{e\}$ , we have

$${\pi }_1(X)\cong \mathbb{Z}\oplus \{e\}\cong \mathbb{Z}.\text{\hspace{1em}(4.13)}$$

来源与版本

正文：英文 · OCR 机器稿 · 待校对

核对状态：OCR 机器稿 · 待校对

原书 PDF

来源版本：second-edition-local

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