The Covering Transformation Theorem - Fund. Group of the Circle

The Covering Transformation Theorem - Fundamental Group of the Circle

Recall from The Covering Transformation Theorem for Finding Fundamental Groups page that if $X$ is a topological space and $(\tilde{X}, p)$ is a universal cover of $X$ then:

\begin{align} \quad A(\tilde{X}) \cong \pi_1(X, x) \end{align}

(Where $A(\tilde{X})$ denotes the set of covering transformations of $\tilde{X}$.)

We will now look at applying this result to find the fundamental group of the circle, $S^1$.

Recall that if $X = S^1$ then $\tilde{X} = \mathbb{R}$ is the universal cover of $X$ with the covering map $p : \tilde{X} \to X$ defined by:

\begin{align} \quad p(x) = (\cos 2\pi x, \sin 2\pi x) \end{align}

In a sense, we wrap $\mathbb{R}$ around on itself a countable number of times. For each $n \in \mathbb{Z}$ let $f_n : \mathbb{R} \to \mathbb{R}$ be defined for all $x \in \mathbb{R}$ by:

\begin{align} \quad f_n(x) = x + n \end{align}

Observe that for each $n \in \mathbb{N}$, $f_n : \mathbb{R} \to \mathbb{R}$ is a covering transformation of $\mathbb{R}$. Also note that $A(\tilde{X})$ is generated by $f_1$, and so:

\begin{align} \quad A(\tilde{X}) \cong \mathbb{Z} \cong \pi_1(X, x) = \pi_1(S^1, x) \end{align}

So we have finally proven that the fundamental group of the circle is isomorphic to $\mathbb{Z}$.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License