# 8.1 ${\pi}_{1}({S}_{1})$

In this section^{}, our goal is to show that ${\pi}_{1}({\mathbb{S}}^{1})=\mathbb{Z}$.
In fact, we will show that the loop space^{} $\mathrm{\Omega}({\mathbb{S}}^{1})$ is equivalent^{} to $\mathbb{Z}$.
This is a stronger statement, because ${\pi}_{1}({\mathbb{S}}^{1})=\parallel \mathrm{\Omega}({\mathbb{S}}^{1}){\parallel}_{0}$ by
definition; so if $\mathrm{\Omega}({\mathbb{S}}^{1})=\mathbb{Z}$, then $\parallel \mathrm{\Omega}({\mathbb{S}}^{1}){\parallel}_{0}=\parallel \mathbb{Z}{\parallel}_{0}$ by congruence^{}, and
$\mathbb{Z}$ is a set by definition (being a set-quotient; see \autorefdefn-Z,\autorefZ-quotient-by-canonical-representatives), so $\parallel \mathbb{Z}{\parallel}_{0}=\mathbb{Z}$.
Moreover, knowing that $\mathrm{\Omega}({\mathbb{S}}^{1})$ is a set will imply that ${\pi}_{n}({\mathbb{S}}^{1})$ is trivial for $n>1$, so we will actually have calculated *all* the homotopy groups of ${\mathbb{S}}^{1}$.

Title | 8.1 ${\pi}_{1}({S}_{1})$ |

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