Difference between revisions of "Circle"

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

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

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

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

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

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

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

• $\frac{\mathbb{R} }{\mathbb{Z} }$ is compact as it is the image of a compact set, namely $[0,1]$ under $q$
• $\mathbb{S}^1$ is Hausdorff since it is a metric space and every metric space is Hausdorff.
• $f$ is surjective, so as $f=\bar{f}\circ q$ and $q$ is surjective, $\bar{f}$ must be too.
  • Otherwise there'd be things $f$ maps to that $\bar{f}\circ q$ may not - contradicting the diagrams commute
• $\bar{f}$ is injective
  • To be injective $\bar{f}([t_1])=\bar{f}([t_2])\implies[t_1]=[t_2]$
    • Showing that $\bar{f}$ is well defined

      Given $a,b\in [t]$ we know $a\sim b$ as $[t]$ is an Equivalence class

      So this means $f(a)=f(b)$ because that's how we defined 'equivalent'

      Thus $\forall a\in [t][\bar{f}([t])=f(a)]$ - so we can defined $\bar{f}$ unambiguously!

    • Using this we see $\bar{f}([t_1])=f(t_1)$ (choosing $t_1$ as the representative of $[t_1]$) and

      $\bar{f}([t_2])=f(t_2)$, so we have $f(t_1)=f(t_2)$ so $t_1\sim t_2$

      • Now we know $t_1\in[t_1]\cap[t_2]$ and $t_2\in[t_1]cap[t_2]$

      As cosets are either disjoint or equal, and they're not disjoint! (we know $t_1$ is in the intersection even if $t_1=t_2$)

      so are equal - thus $\bar{f}$ is injective.

• Thus $\bar{f}$ is a bijection

Using the Compact-to-Hausdorff theorem we conclude $\bar{f}$ is a homeomorphism