# 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 ${\Omega(\mathbb{S}^{1})}$ is equivalent to $\mathbb{Z}$. This is a stronger statement, because $\pi_{1}(\mathbb{S}^{1})=\mathopen{}\left\|\Omega(\mathbb{S}^{1})\right\|_{0}% \mathclose{}$ by definition; so if $\Omega(\mathbb{S}^{1})=\mathbb{Z}$, then $\mathopen{}\left\|\Omega(\mathbb{S}^{1})\right\|_{0}\mathclose{}=\mathopen{}% \left\|\mathbb{Z}\right\|_{0}\mathclose{}$ by congruence, and $\mathbb{Z}$ is a set by definition (being a set-quotient; see \autorefdefn-Z,\autorefZ-quotient-by-canonical-representatives), so $\mathopen{}\left\|\mathbb{Z}\right\|_{0}\mathclose{}=\mathbb{Z}$. Moreover, knowing that ${\Omega(\mathbb{S}^{1})}$ 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 $\pi_{1}(S_{1})$ \metatable