View source for Circle

{{Extra Maths}}
==Definition==
A circle is usually defined by {{M|1=\mathcal{S}^1=\Big\{(x,y)\in\mathbb{R}^2{{!}}d\Big((0,0),(x,y)\Big)=1 \Big\} }}

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

===The circle as a quotient space===
{{Begin Theorem}}
Theorem: The circle {{M|\mathbb{S}^1}} is [[Homeomorphism|homeomorphic]] to {{M|\frac{\mathbb{R} }{\mathbb{Z} } }}
{{Begin Proof}}
Using the map above, we see that this just wraps the real line around the circle over and over again, specifically {{M|1=f(t_1)=f(t_2)\iff t_1-t_2\in\mathbb{S} }}, this suggests an [[Equivalence relation]].

Using a bit of abstract algebra it is not hard to see that the [[Equivalence class|equivalence classes]] are exactly the [[Coset|cosets]] of {{M|\mathbb{Z} }} in {{M|\mathbb{R} }}. So it is no problem to write {{M|1=\tfrac{\mathbb{R} }{\sim}=\tfrac{\mathbb{R} }{\mathbb{Z} } }}


Using [[Passing to the quotient]] we see that {{M|\exists\bar{f} }} that makes the diagram below commute '''if and only if''' {{M|1=t_1\sim t_2\implies f(t_1)=f(t_2)}}

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


[math]
\begin{CD}
R {{CD Hoz Eq}} R \\
{{CD Down Arrow|q}} {{CD Down Arrow||f}} \\
\frac{\mathbb{R}}{\mathbb{Z}} {{CD Right Arrow||\bar{f} }} \mathbb{S}^1
\end{CD}
[/math] {{Triangle Wanted}}

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


If {{M|\bar{f} }} is a homeomorphism the result is shown. 

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

Using the [[Compact-to-Hausdorff theorem]] we conclude {{M|\bar{f} }} is a '''homeomorphism'''
{{End Proof}}
{{End Theorem}}

==See also==
* [[Sphere]]

{{Definition|Manifolds|Topology}}