---- > [!theorem] Theorem. ([[the fundamental group of the circle is infinite cyclic]]) > Let $u: I \to \mathbb{S}^{1}$ be the [[parameterized curve|loop]] given by $t \mapsto e^{2\pi i t}$, which is based at $1 \in \mathbb{S}^{1} \subset \mathbb{C}$. Then there is a [[group isomorphism]] $\pi_{1}(\mathbb{S}^{1}, 1) \cong (\mathbb{Z}, +)$ sending $[u]$ to $1$. > [!proof]- Proof. ([[the fundamental group of the circle is infinite cyclic]]) > Let $p:\mathbb{R} \to \mathbb{S}^{1}$ be the [[covering space|covering map]] of $\mathbb{S}^{1}$ defined [[covering space|here]], let $e_{0}=0$, and let $b_{0}=p(e_{0})=(1,0)$. Then $p ^{-1}(b_{0})=\mathbb{Z}$. Since $\mathbb{R}$ is [[simply connected]], the [[lifting correspondence derived from a covering map|lifting correspondence]] $\phi:\pi_{1}(\mathbb{S}^{1}, b_{0})\to\mathbb{Z}$ [[path-connected covering space yields surjective lifting correspondence|is]] a [[bijection]]. Thus all we must do is show $\phi$ is a [[group homomorphism|homomorphism]]. > Given $[f],[g] \in \pi_{1}(\mathbb{S}^{1},b_{0})$, let $\tilde{f}$ and $\tilde{g}$ be their respective [[lifting|liftings]] to [[parameterized curve|paths]] on $\mathbb{R}$ beginning at $0=e_{0}$. Let $n=\tilde{f}(1)$ and $m=\tilde{g}(1)$; then $\phi([f])=n$ and $\phi([g])=m$, by definition. Let $\tilde{\tilde{g}}$ be the 'translate' [[parameterized curve]] $\tilde{\tilde{g}}(s)=n+\tilde{g}(s)$ on $\mathbb{R}$. Because $p(n+x)=p(x)$ for all $x \in \mathbb{R}$, $\tilde{\tilde{g}}$ is a [[lifting]] of $g$ beginning at $n$. Then the product $\tilde{f} * \tilde{\tilde{g}}$ is defined, and it is the lifting of $f * g$ that begins at $0$, as you can check. The end point of this [[parameterized curve]] is $\tilde{\tilde{g}}(1)=n+m$. Then by definition, $\phi([f] * [g])=n+m=\phi([f])+\phi([g]).$ Recall from the examples in [[covering space|covering map]] that there is a [[covering space|covering map]] $\mathbb{R}\to \mathbb{S}^{1}$ via $p(t):=e^{2\pi i t}$. Since $\mathbb{R}$ is [[simply connected]], this is a [[universal cover]]. Put $x_{0}:=1$. As a set we have $p ^{-1}(x_{0})=\mathbb{Z} \subset \mathbb{R}$; choose as [[pointed set|basepoint]] $0 \in p ^{-1}(x_{0})$. [[group action|Acting]] via the [[monodromy action]] on this basepoint gives, via [[the monodromy correspondence]][^1]), a [[bijection]] $\ell: \pi_{1}(\mathbb{S}^{1}, x_{0}) \xrightarrow{\sim}p ^{-1}(x_{0})=\mathbb{Z}.$ Recall from corollary in [[the monodromy correspondence]] that $\pi_{1}(\mathbb{S}^{1}, x_{0})$ induces a multiplication law on $p ^{-1}(x_{0})$. Following the 3-step template provided there, we see in this case that the product of $m,n \in \mathbb{Z}$ is obtained by: >1. choosing a path $\tilde{\gamma}:I \to \mathbb{R}$ from $0$ to $m$ ([[path homotopies lift uniquely under covering maps|unique up to homotopy]]), such as the [[parameterized curve]] $\tilde{u}_{m}: I \to \mathbb{R}$ given by $\tilde{u}_{m}(t)=mt$. >2. letting $\gamma: I \to \mathbb{R}$ be the [[lifting|lift]] of the [[parameterized curve|loop]] $p \circ \tilde{\gamma}$ (in this case given by $p \circ \tilde{u}_{m}(t)=e^{2\pi i m t}$), starting at $n$. By inspection, this [[lifting|lift]] is $n+\tilde{u}_{m}(t)=n+mt$. >3. now $\ell( \ell ^{-1}(m) * \ell ^{-1}(n) )$ equals the endpoint of $\gamma$, i.e., $n+m$. > This proves the result. ---- #### [^1]: Or in taking [[path-connected covering space yields surjective lifting correspondence|this result]] after defining [[lifting correspondence derived from a covering map|lifting correspondence]] in terms of $y_{0}=0$. It is the same exact thing, just different verbiage. ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM outgoing([[]]) FLATTEN file.tags GROUP BY file.tags as Tag