# Circle

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## Definition

A circle is usually defined by [ilmath]\mathcal{S}^1=\Big\{(x,y)\in\mathbb{R}^2|d\Big((0,0),(x,y)\Big)=1 \Big\}[/ilmath]

## Topological perspective

The map [ilmath]f:\mathbb{R}\rightarrow\mathbb{S}^1[/ilmath] given by [ilmath]f:t\mapsto e^{2\pi jt} [/ilmath] is significant. As it makes [ilmath]\mathbb{R} [/ilmath] a covering space of [ilmath]\mathbb{S}^1[/ilmath]

### The circle as a quotient space

Theorem: The circle [ilmath]\mathbb{S}^1[/ilmath] is homeomorphic to [ilmath]\frac{\mathbb{R} }{\mathbb{Z} } [/ilmath]

Using the map above, we see that this just wraps the real line around the circle over and over again, specifically [ilmath]f(t_1)=f(t_2)\iff t_1-t_2\in\mathbb{S}[/ilmath], this suggests an Equivalence relation.

Using a bit of abstract algebra it is not hard to see that the equivalence classes are exactly the cosets of [ilmath]\mathbb{Z} [/ilmath] in [ilmath]\mathbb{R} [/ilmath]. So it is no problem to write [ilmath]\tfrac{\mathbb{R} }{\sim}=\tfrac{\mathbb{R} }{\mathbb{Z} }[/ilmath]

Using Passing to the quotient we see that [ilmath]\exists\bar{f} [/ilmath] that makes the diagram below commute if and only if [ilmath]t_1\sim t_2\implies f(t_1)=f(t_2)[/ilmath]

$\begin{xy} \xymatrix{ {\mathbb{R}} \ar[d] \ar[dr] &\\ {\frac{\mathbb{R}}{\mathbb{Z}}} \ar[r]_{\bar{f}} & {\mathbb{S}^1} } \end{xy}$

$\begin{CD} R @= R \\ @V q V V @V Vf V \\ \frac{\mathbb{R}}{\mathbb{Z}} @> >\bar{f} > \mathbb{S}^1 \end{CD}$ (Triangle diagram wanted)

(Where [ilmath]\bar{f}:\frac{\mathbb{R} }{\mathbb{Z} }\rightarrow{\mathbb{S}^1} [/ilmath] is given by [ilmath]\bar{f}:[t]\rightarrow f(t)[/ilmath])

If [ilmath]\bar{f} [/ilmath] is a homeomorphism the result is shown.

• [ilmath]\frac{\mathbb{R} }{\mathbb{Z} } [/ilmath] is compact as it is the image of a compact set, namely [ilmath][0,1][/ilmath] under [ilmath]q[/ilmath]
• [ilmath]\mathbb{S}^1[/ilmath] is Hausdorff since it is a metric space and every metric space is Hausdorff.
• [ilmath]f[/ilmath] is surjective, so as [ilmath]f=\bar{f}\circ q[/ilmath] and [ilmath]q[/ilmath] is surjective, [ilmath]\bar{f} [/ilmath] must be too.
• Otherwise there'd be things [ilmath]f[/ilmath] maps to that [ilmath]\bar{f}\circ q[/ilmath] may not - contradicting the diagrams commute
• [ilmath]\bar{f} [/ilmath] is injective
• To be injective [ilmath]\bar{f}([t_1])=\bar{f}([t_2])\implies[t_1]=[t_2][/ilmath]
• Showing that [ilmath]\bar{f} [/ilmath] is well defined
Given [ilmath]a,b\in [t][/ilmath] we know [ilmath]a\sim b[/ilmath] as [ilmath][t][/ilmath] is an Equivalence class
So this means [ilmath]f(a)=f(b)[/ilmath] because that's how we defined 'equivalent'
Thus [ilmath]\forall a\in [t][\bar{f}([t])=f(a)][/ilmath] - so we can defined [ilmath]\bar{f} [/ilmath] unambiguously!
• Using this we see [ilmath]\bar{f}([t_1])=f(t_1)[/ilmath] (choosing [ilmath]t_1[/ilmath] as the representative of [ilmath][t_1][/ilmath]) and
[ilmath]\bar{f}([t_2])=f(t_2)[/ilmath], so we have [ilmath]f(t_1)=f(t_2)[/ilmath] so [ilmath]t_1\sim t_2[/ilmath]
• Now we know [ilmath]t_1\in[t_1]\cap[t_2] [/ilmath] and [ilmath]t_2\in[t_1]cap[t_2] [/ilmath]
As cosets are either disjoint or equal, and they're not disjoint! (we know [ilmath]t_1[/ilmath] is in the intersection even if [ilmath]t_1=t_2[/ilmath])
so are equal - thus [ilmath]\bar{f} [/ilmath] is injective.
• Thus [ilmath]\bar{f} [/ilmath] is a bijection

Using the Compact-to-Hausdorff theorem we conclude [ilmath]\bar{f} [/ilmath] is a homeomorphism