The circle as $S^1=\mathbb{R}\mathrm{/}\mathbb{Z}$ with basepoint $[0]$

Use the canonical copy of $\mathbb{Z}$ inside $\mathbb{R}$ fixed in Integer part: for every real $x$ there is exactly one integer $m$ with $m\le x<m+1$. For $x,y\in \mathbb{R}$, put

$x\sim y⟺x-y\in \mathbb{Z}\mathrm{.}$

This is an equivalence relation. Indeed, $x-x=0\in \mathbb{Z}$; if $x-y\in \mathbb{Z}$, then $y-x=-(x-y)\in \mathbb{Z}$; and if $x-y,y-z\in \mathbb{Z}$, then $x-z=(x-y)+(y-z)\in \mathbb{Z}$. The closure facts used here are part of the additive-group structure supplied by The integers form a commutative ring, and the quotient-set construction is that of The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection.

Let $[x]$ denote the equivalence class of $x$. Let $p:\mathbb{R}\to \mathbb{R}\mathrm{/}\mathbb{Z}$ be the canonical projection, $p(x)=[x]$. Thus

$p(x)=p(y)⟺x-y\in \mathbb{Z}\mathrm{.}$

Let $\mathbb{R}\mathrm{/}\mathbb{Z}$ carry the quotient topology induced by $p$. The circle is $S^1:=\mathbb{R}\mathrm{/}\mathbb{Z}$ with the quotient topology induced by $p(x)=[x]$ and basepoint $[0]$; moreover $p^{-1}([0])=\mathbb{Z}$ and $p(x+n)=p(x)$ for every real $x$ and integer $n$.

The last assertions follow directly from the displayed fibre criterion: $p(x)=[0]$ exactly when $x\in \mathbb{Z}$, while $(x+n)-x=n\in \mathbb{Z}$.