Surjective Function

**ejgmath** · May 13th 2010, 08:01 PM

Let $X=\mathbb{R}$ then define an equivalence relation $\sim$ on $X$ s.t.

$x\sim y$ if and only if $x-y\in\mathbb{Z}$

Show that $X/\sim\cong S^1$

So denoting the elements of $X/\sim$ as $[t]$

The function

$f([t])=\exp^{2\pi ti}$ defines a homemorphism.

$f([t])$ is continuous since $\exp^{2\pi ti}=cos(2\pi t)+isin(2\pi t)$ which is the sum of continous functions.

Letting $f([x])=f([y])\Rightarrow\exp^{2\pi xi}=\exp^{2\pi yi}\Rightarrow2\pi xi=2\pi yi\Rightarrow x=y$ so injective.

Now $f([t])=\exp^{2\pi ti}=z$ for $z\in\mathbb{C}$ s.t. $|z|=1$

Then $t=\frac{-i}{2\pi}log(z)=\frac{-i}{2\pi}(log|z|+iArg(z))$

Since $|z|=1$ we have $log|z|=0$ so $t=\frac{1}{2\pi}Arg(z)$ which is in $X/\sim$ so surjective.

Therefore a bijection.

Not sure how to show $f^{-1}$ continuous?

Is this correct? Any input would be great. Thanks

**Drexel28** · May 13th 2010, 08:11 PM

$f([t])=\exp^{2\pi ti}$ defines a homemorphism.

True. But the mapping is only well defined (independent of representative from the equivalence class) because $e^{2\pi i t}$ is periodic over the integers.

$f([t])$ is continuous (trivial)

Letting $f([x])=f([y])\Rightarrow\exp^{2\pi xi}=\exp^{2\pi yi}\Rightarrow2\pi xi=2\pi yi\Rightarrow x=y$ which implies injective.

Saying it is slightly more subtle, the above is true ONLY because of the construction of $X/\sim$ . Say more.

And yes it is surjective. But, why do it this way? Really you aren't done, you haven't proven the inverse function is continuous.

Have you proven that $X/\sim$ is Hausdorff? Then, reverse the direction of your mapping and you only have to show that your function is continuous (you already proved it was bijective) and thus you will have a continuous bijection from a compact space into a Hausdorff space and thus automatically a homeomorphism.

That said, you've already done everything else. Might as well finish it.

**Drexel28** · May 13th 2010, 08:22 PM

Originally Posted by ejgmath

$f([t])$ is continuous since $\exp^{2\pi ti}=cos(2\pi t)+isin(2\pi t)$ which is the sum of continous functions.

Just a remark in taste. Although what you said is true I think it is personally more enlightening to realize that $\mathbb{C}\approx\mathbb{R}^2$ and so you're mapping will be continuous if and only if $\alpha:X/\sim\to\mathbb{R}^2:[t]\mapsto \left(\cos(2\pi [t]),\sin(2\pi[t])\right)$ from where continuity is much more clear since the coordinate functions $\pi\alpha:X/\sim\to\mathbb{R}:[t]\mapsto\cos(2\pi[t])$ are continuous. This isn't altogether clear (at least to me) but remember the defining characteristic of quotient maps. Namely, $\pi\alpha$ is continuous if and only if so is $\pi\circ\alpha\circ\xi:\mathbb{R}\to\mathbb{R}:t\m apsto \cos(2\pi[t])=\cos(2\pi t)$ . There, now it is OBVIOUS that it's continuous

EDIT: $\xi:\mathbb{R}\to \mathbb{R}/\mathbb{Z}:t\mapsto [t]$ was meant to be the quotient map.

P.S. Interestingly enough we can go further. If we consider $\mathbb{S}^1$ to be both a topological space with the usual topology and a group under complex multiplication, we give $\mathbb{R}/\mathbb{Z}$ the quotient topology and the group structure as a quotient group, and finally give $\text{SO}(2)]$ the normal topology and matrix multiplication it is true that $\mathbb{R}/\mathbb{Z}\overset{\text{T.G.}}{\cong}\mathbb{S}^1 \overset{\text{T.G.}}{\cong}\text{SO}(2)$ where the T.G. means topological group isomorphism.

**ejgmath** · May 13th 2010, 09:16 PM

Thanks for the responses, after writing my post I realised I had missed out the inverse function continuity.

Just a quick question, if I let $f^{-1}(z)=\frac{1}{2\pi}Arg(z)$ that would clearly create the equivalence class for any $t$ that generates $z$ .

But what would you say about it's continuity? Is $Arg(z)$ continuous by definition?

**Drexel28** · May 13th 2010, 09:24 PM

Originally Posted by ejgmath

Thanks for the responses, after writing my post I realised I had missed out the inverse function continuity.

Just a quick question, if I let $f^{-1}(z)=\frac{1}{2\pi}Arg(z)$ that would clearly create the equivalence class for any $t$ that generates $z$ .

But what would you say about it's continuity? Is $Arg(z)$ continuous by definition?

Honestly, I'm not too sure if something gets screwed up. I'll have to check. Have you thought about proving that $X/\sim$ is compact (it's not too hard) and then you're homeomorphism is immediate.

**ejgmath** · May 13th 2010, 09:27 PM

Yeh, I did consider that but the (revision) question is specifically looking for the whole continuous bijection business.

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