📂Topological Data Analysis

The Fundamental group of a Circle is Isomorphic to the Integer group

Theorem

The fundamental group $\pi_{1} \left( S^{1}, 1 \right)$ of the unit circle $S^{1}$ is isomorphic to the integer group $\mathbb{Z}$. $$\pi_{1} \left( S^{1}, 1 \right) \simeq \mathbb{Z}$$ Here, $(1,0) \in \mathbb{R}^{2}$ is also simply denoted as $1$.

Explanation

Application

Studying homotopy, one of the biggest and most important results that can be obtained is that the fundamental group of $S^{1}$ is the integer group. Known applications include:

• A topological proof of the Fundamental Theorem of Algebra
• Brouwer Fixed Point Theorem
• Borsuk-Ulam Theorem

However, the significance of this theorem lies in its own right, regardless of these applications. From an algebraic topology perspective, the idea of studying spaces not just with points but through loops makes $S^{1}$ an essentially primary object to consider.

Intuitive Meaning

Basically, “the fundamental group of a circle being isomorphic to the integers” means nothing more than the number of times (n) a loop wraps around the circle in either the counterclockwise (+) or clockwise (-) direction, laying the algebraic groundwork to tackle some of the tougher areas of topology.

The Fundamental group of a Multi-Dimensional Sphere is Trivial

$$\pi_{1} \left( S^{n} \right) \simeq 0 \qquad , n \ge 2$$ For $n \ge 2$, the fundamental group of the $n$-sphere $S^{n}$ is the Trivial group.

Intuitively, imagining $n = 2$ when $S^{2}$, all loops can inevitably be shrunk down to a point across the surface, as shown in the figure above. For a formal proof, refer to Hatcher¹.

Proof ²

Part 1. Definition of the function $\varphi$

By the definition of the fundamental group, all loops $[f] \in \pi_{1} \left( S^{1}, 1 \right)$ must start and end at $1$. Let $\deg_{+} (f)$ be the number of times $[f]$ makes a complete counterclockwise rotation to return to $1$, and let $\deg_{-} (f)$ be the number of clockwise rotations when $\varphi : \pi_{1} \left( S^{1}, 1 \right) \to \mathbb{Z}$ is defined as follows. $$\varphi \left( \left[ f \right] \right) := \deg_{+} (f) - \deg_{-} (f)$$

Monodromy Theorem: Given two path $f_{0} \simeq f_{1}$ equivalent at the starting point $1 := (1,0)$ on $1$-sphere $S^{1}$, if their respective lifts satisfy $\widetilde{f}_{0}, \widetilde{f}_{1}$ equals $\widetilde{f}_{0} (0) = \widetilde{f}_{1} (0)$, then $\widetilde{f}_{0} (1) = \widetilde{f}_{1} (1)$ holds.

If $\widetilde{f} (0) = 0$, then by the Monodromy Theorem, there is a unique $\widetilde{f} : I \to \mathbb{R}$ such that $\varphi \left( [f] \right) = \widetilde{f} (1)$, and $\varphi$ is a function. Next, it is shown that $\varphi$ is an isomorphism.

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Part 2. $\varphi$ is a homomorphism

Let $l_{a} (f)$ be the lift of $f$ starting from $a \in p^{-1} \left( f(0) \right)$, then $l_{0} (f) = \widetilde{f}$ holds, and for all $a \in \mathbb{R}$, $$\left( l_{a} (f) \right) (t) = \widetilde{f} (t) + a$$ it follows. Here, if $b \in \mathbb{R}$ is set to $b := \widetilde{f} (t) + a$, then trivially the following applies based on path operation $\ast$. $$l_{a} \left( f \ast g \right) = l_{a} (f) \ast l_{b} (g) \qquad \cdots 🤔$$ Thus, when $a = 0$, $b = \widetilde{f} (1)$ holds, and for $[f] , [g] \in \pi_{1} \left( S^{1}, 1 \right)$, $$\begin{align*} & \varphi \left( [f] [g] \right) \\ =& \varphi \left( \left[ f \ast g \right] \right) \\ =& \left( \widetilde{f \ast g} \right) (1) \\ =& l_{0} \left( f \ast g \right) (1) \\ =& \left( l_{0} (f) \ast l_{b} (g) \right) (1) \qquad \because 🤔 \\ =& l_{b} (g) (1) \\ =& b + \widetilde{g} (1) \\ =& \widetilde{f} (1) + \widetilde{g} (1) \\ =& \varphi \left( [f] \right) + \varphi \left( [g] \right) \end{align*}$$ $\varphi$ preserves the operation in $\pi_{1} \left( S^{1}, 1 \right)$ in $\mathbb{Z}$, making it a homomorphism.

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Part 3. $\varphi$ is surjective

Given $n \in \mathbb{Z}$, by defining $\omega_{n} : I \to \mathbb{R}$ as $\omega_{n} (t) := n t$, $$p \circ \omega_{n} : I \to \mathbb{R} \to S^{1}$$ it becomes a loop with $1 \in S^{1}$ as the base point. By definition, $\omega_{n}$ is a lift of $\omega_{n} (0) = 0$ while being $p \circ \omega_{n}$, and $$\varphi \left( [ p \circ \omega_{n} ] \right) = \varphi \left( p \circ \omega_{n} \right) = \omega_{n} (1) = n$$ holds. Thus, for every $n \in \mathbb{Z}$, there exists $[ p \circ \omega_{n} ] \in \pi_{1} \left( S^{1}, 1 \right)$ satisfying $\varphi \left( [ p \circ \omega_{n} ] \right) = n$, making $\varphi$ surjective.

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Part 4. $\varphi$ is injective

Homotopy Lifting Property: A continuous function $F : I^{2} \to S^{1}$ has a lift $\widetilde{F} : I^{2} \to \mathbb{R}$. Specifically, for given $x_{0} \in S^{1}$ and $\widetilde{x}_{0} \in p^{-1} \left( x_{0} \right)$, there exists a unique $\widetilde{F}$ such that $\widetilde{F} \left( 0 , 0 \right) = \widetilde{x}_{0}$.

Assume $\varphi \left( [f] \right) = 0$. Here, remember that $0 \in \mathbb{Z}$ is not just $0$, but $0$ as the additive identity of integers. This means the starting and ending points of the lift $\widetilde{f} : I \to \mathbb{R}$ of $f$ are the same, and due to the uniqueness of the lift, $$\widetilde{f} (0) = \widetilde{f} (1) = 0$$ according to the Homotopy Lifting Property, $$\begin{align*} F (0,t) =& \widetilde{f} (t) \\ F (1,t) =& 0 \\ F (s,0) = F (s,1) =& 0 \end{align*}$$ $F : I^{2} \to \mathbb{R}$ satisfying the condition exists. Naturally, applying the covering (projection) $p$ to $$p \circ F : I^{2} \to \mathbb{R} \to S^{1}$$ satisfies $$\begin{align*} p \circ F (0,t) =& f (t) \\ p \circ F (1,t) =& 1 \\ p \circ F (s,0) = p \circ F (s,1) =& 1 \end{align*}$$, meaning $[f] \in \pi_{1} \left( S^{1}, 1 \right)$ is the constant path $[f] = [c_{1}]$ of $\pi_{1} \left( S^{1}, 1 \right)$, thus $\ker \varphi = \left\{ \left[ c_{1} \right] \right\}$.

Properties of the Kernel of a Homomorphism:

• [3]: $\ker \phi = \left\{ e \right\}$ $\iff$ $\phi$ is injective.

Lastly, by the properties of the kernel, $\varphi$ is injective.

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Meanwhile, Hatcher defines the domain and range conversely, proving that $\phi : \mathbb{Z} \to \pi_{1} \left( S^{1}, 1 \right)$ is an isomorphism, though essentially arriving at the same conclusion but aiming to prove it without mentioning the Monodromy Theorem etc¹.