The Fundamental Group

Group Structure on Loops

Fundamental Group

The fundamental group \newcommand{\R}{\mathbb{R}}\newcommand{\Z}{\mathbb{Z}}\pi_{1}(X,x_{0}) of a topological space $X$ at a base point $x_0$ is the group of homotopy classes of loops based at $x_0$, with the multiplication of concatenation of loops, inverse being the reversed loop, and identity being the constant loop at $x_0$.

Lemma

In $\mathbb{R}$, any two paths are homotopic if and only if they have the same endpoints.

Proof For any two paths $f$ and $g$ with same endpoints, $H(t,s)=t\cdot f(s)+(1-t)\cdot g(s)$ gives a valid homotopy. $\square$

Proposition

${\pi }_1(S^1,1)\cong \mathbb{Z}$.

Proof The ideal is that $S^1$ looks locally like $\mathbb{R}$. Define $p:\mathbb{R}\to S^1$, $x\mapsto e^{2\pi ix}$, paths $w_n:I\to S^1$, $s\mapsto e^{2\pi ins}$, and ${\tilde{w}}_n:I\to \mathbb{R}$, $s\mapsto ns$. Note that $p\circ {\tilde{w}}_n=w_n$. We claim that $\Phi :\mathbb{Z}\to {\pi }_1(S^1,1)$ sending integer $n$ to the homotopy class of $w_n$ forms the desired isomorphism. First show that it is a group homomorphism, i.e. $w_{m+n}\simeq w_m\cdot w_n$. Define ${\tau }_m:\mathbb{R}\to \mathbb{R}$, $x\mapsto x+m$. Observe that $(p\circ ({\tau }_m\circ {\tilde{w}}_n))(x)=e^{2\pi i(nx+m)}=e^{2\pi inx}=w_n(x),$so $p\circ ({\tau }_m\circ {\tilde{w}}_n)=w_n$. Note that both ${\tilde{w}}_{m+n}$ and ${\tilde{w}}_m\cdot ({\tau }_m\circ {\tilde{w}}_n)$ are paths in $\mathbb{R}$ from $0$ to $m+n$, hence there exists a homotopy ${\tilde{f}}_t$ from ${\tilde{w}}_m\cdot ({\tau }_m\circ {\tilde{w}}_n)$ to ${\tilde{w}}_{m+n}$. Now let $f_t:=p\circ {\tilde{f}}_t$, then $f_t$ is a homotopy from $p\circ ({\tilde{w}}_m\cdot ({\tau }_m\circ {\tilde{w}}_n))=w_m\cdot w_n$ to $p\circ {\tilde{w}}_{m+n}=w_{m+n}$. This shows that $\Phi$ is a homomorphism. Now we show the surjectivity. Fixing some loop $f:I\to S^1$. Define $S_{+}:=S^1\setminus \{1\}$ and $S_{-}:=S^1\setminus \{-1\}$. We can partition the preimage of $p$ as following: \begin{align} p^{-1}(S_{+})=\bigsqcup_{n\in\Z}\tilde{V}_{n},&\quad \tilde{V}_{n}=(n,n+1), \\ p^{-1}(S_{-})=\bigsqcup_{n\in\Z}\tilde{U}_{n},&\quad \tilde{U}_{n}=(n-1/2,n+1/2). \end{align}Then the restrictions of $p$, say $p{|}_{{\tilde{U}}_n}:{\tilde{U}}_n\to S_{-}$ and $p{|}_{{\tilde{V}}_n}:{\tilde{V}}_n\to S_{+}$ are homeomorphism, with inverses ${\alpha }_n:S_{-}\to {\tilde{U}}_n$ and ${\beta }_n:S_{+}\to {\tilde{V}}_n$ respectively. There is a partition of $I$, $0=s_0<s_1<\cdots <s_k=1$ with each $f([s_{j-1},s_j])$ either a subset of $S_{+}$ or a subset of $S_{-}$. Now define $\tilde{f}:I\to \mathbb{R}$ satisfying $\tilde{f}(0)=0$, and then inductively on $[s_{j-1},s_j]$ by $\tilde{f}{|}_{[s_{j-1},s_j]}:=\{\begin{array}{l}{\alpha }_n\circ f{|}_{[s_{j-1},s_j]},\text{if }f([s_{j-1},j])\subset S_{-},\\ {\beta }_l\circ f{|}_{[s_{j-1},s_j]},\text{if }f([s_{j-1},j])\subset S_{+},\end{array}$where $n$ and $l$ are the unique integer for which $\tilde{f}(s_{j-1})\in {\tilde{U}}_n$ or $\tilde{f}(s_{j-1})\in {\tilde{V}}_l$. This gives a path $\tilde{f}:I\to \mathbb{R}$ starting at $0$ and ending at some $k\in p^{-1}(1)=\mathbb{Z}$. Thus $\tilde{f}\simeq {\tilde{w}}_k$, hence, $f\simeq w_k$ by applying $p$ to the both sides, and so $\Phi (k)=[f]$. Finally, we show that $\Phi$ is injective. Suppose $\Phi (n)=\Phi (m)$, then $w_n\simeq w_m$, there exists a homotopy $H:I\times I\to S^1$ from $w_n$ to $w_m$. Similarly, there exists a partition of $I\times I$, $0=s_0<s_1<\cdots <s_a=1$ and $0=t_0<t_1<\cdots <t_b=1$, such that $H([s_{j-1},s_j]\times [t_{k-1},t_k])$ is either a subset of $S_{+}$ or a subset of $S_{-}$. Now define $\tilde{H}:I\times I\to \mathbb{R}$ similarly as above, which gives a homotopy from ${\tilde{w}}_n$ to ${\tilde{w}}_m$, hence $\Phi (n)=\Phi (m)$ implies $n=m$. Thus $\Phi$ is injective. $\square$

Remark

In fact, the distributive law holds in general: Suppose $\phi :X\to Y$ is continuous, and $f,g:I\to X$ are paths, then $\phi \circ (f\cdot g)=(\phi \circ f)\cdot (\phi \circ g).$

Change of Basepoints

Proposition

Suppose $h:I\to X$ is a path from $x_0$ to $x_1$, then the map ${\beta }_h:{\pi }_1(X,x_1)\to {\pi }_1(X,x_0)$ defined by ${\beta }_h([f]):=[h\cdot f\cdot h^{-1}]$ is an isomorphism.

Proof We first check that it is well-defined. Suppose $f_0\simeq f_1$ are homotopic paths of $[f]\in {\pi }_1(X,x_1)$ with homotopy $f_t$. Then $h\cdot f_0\cdot h^{-1}\simeq h\cdot f_1\cdot h^{-1}$ via $h\cdot f_t\cdot h^{-1}$. Secondly, it is a homomorphism because ${\beta }_h([f][g])={\beta }_h([f\cdot g])=[h\cdot f\cdot g\cdot h^{-1}]=[h\cdot f\cdot h^{-1}\cdot h\cdot g\cdot h^{-1}]={\beta }_h([f]){\beta }_h([g]).$Moreover, ${\beta }_{h^{-1}}$ is a inverse of ${\beta }_h$ (here $h^{-1}$ means the reversed path), thus ${\beta }_h$ is an isomorphism. $\square$

Remark

Note that there is no ambiguity when we defined ${\beta }_h([f]):=[h\cdot f\cdot h^{-1}]$. Because actually $(h\cdot f)\cdot h^{-1}\simeq h\cdot (f\cdot h^{-1})$.

Fundamental groups are powerful in many (other) realms of mathematics.

• It can even be used to prove the fundamental theorem of algebra: Proof For the sake of contradiction, suppose some non-constant polynomial $p(z)=z^n+a_{n-1}{z}^{n-1}+\cdots +a_0$ has no roots. Then for each $r\ge 0$, $f_r(s)=\frac{p(r{e}^{2\pi is})/p(r)}{|p(r{e}^{2\pi is})|/|p(r)|}$is a loop in $S^1$ based at $1$. Now fix some large $R>\max (1,|a_0|+\cdots +|a_{n-1}|)$. Then $f_r$ is a homotopy from $f_0=w_0$ to $f_R$. Let $p_t(z)=z^n+(1-t)(a_{n-1}{z}^{n-1}+\cdots +a_0)$. Define $g_t(s):=\frac{p_t(R{e}^{2\pi is})/p_t(R)}{|p_t(R{e}^{2\pi is})|/|p_t(R)|}.$Note that since $R$ is sufficiently large $p_t(R{e}^{2\pi is})$ is never zero, $g_t$ is a valid homotopy from $f_R$ to $w_n$. Therefore, $w_0=f_0\simeq f_R=g_0\simeq g_1=w_n,$which implies $n=0$ since ${\pi }_1(S^1,1)\cong \mathbb{Z}$. This contradicts the assumption that $p$ is non-constant. $\square$
• We can also show Brouwer’s fixed-point theorem by the fundamental group. It states that any continuous $h:D^2\to D^2$ has a fixed point, where $D^2$ is the closed disk in $\text{C}$. Proof Suppose $h$ has no fixed points. Then we can define $r:D^2\to S^1$ by sending $x\in D^2$ to the intersect point of $S^1$ and the ray from $h(x)$ to $x$. Note that $r(x)=x$ for $x\in S^1$. Consider the inclusion $i:S^1\hookrightarrow D^2$, we have $i\circ w_1\simeq c_1$ via some homotopy $f_t$. Then $w_1=r\circ i\circ w_1\simeq c_1$ via $r\circ f_t$, yielding a contradiction. $\square$