# Surjective Function

#### ejgmath

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

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

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

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

The function

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

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

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

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

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

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

Therefore a bijection.

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

Is this correct? Any input would be great. Thanks

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#### Drexel28

MHF Hall of Honor
$$\displaystyle f([t])=\exp^{2\pi ti}$$ defines a homemorphism.
True. But the mapping is only well defined (independent of representative from the equivalence class) because $$\displaystyle e^{2\pi i t}$$ is periodic over the integers.

$$\displaystyle f([t])$$ is continuous (trivial)
Letting $$\displaystyle f([x])=f([y])\Rightarrow\exp^{2\pi xi}=\exp^{2\pi yi}\Rightarrow2\pi xi=2\pi yi\Rightarrow x=y$$ which implies injective.
Saying it is slightly more subtle, the above is true ONLY because of the construction of $$\displaystyle X/\sim$$. 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 $$\displaystyle X/\sim$$ 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.

• ejgmath

#### Drexel28

MHF Hall of Honor
$$\displaystyle f([t])$$ is continuous since $$\displaystyle \exp^{2\pi ti}=cos(2\pi t)+isin(2\pi t)$$ 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 $$\displaystyle \mathbb{C}\approx\mathbb{R}^2$$ and so you're mapping will be continuous if and only if $$\displaystyle \alpha:X/\sim\to\mathbb{R}^2:[t]\mapsto \left(\cos(2\pi [t]),\sin(2\pi[t])\right)$$ from where continuity is much more clear since the coordinate functions $$\displaystyle \pi\alpha:X/\sim\to\mathbb{R}:[t]\mapsto\cos(2\pi[t])$$ are continuous. This isn't altogether clear (at least to me) but remember the defining characteristic of quotient maps. Namely, $$\displaystyle \pi\alpha$$ is continuous if and only if so is $$\displaystyle \pi\circ\alpha\circ\xi:\mathbb{R}\to\mathbb{R}:t\mapsto \cos(2\pi[t])=\cos(2\pi t)$$. There, now it is OBVIOUS that it's continuous

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

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

• ejgmath

#### 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 $$\displaystyle f^{-1}(z)=\frac{1}{2\pi}Arg(z)$$ that would clearly create the equivalence class for any $$\displaystyle t$$ that generates $$\displaystyle z$$.

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

#### Drexel28

MHF Hall of Honor
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 $$\displaystyle f^{-1}(z)=\frac{1}{2\pi}Arg(z)$$ that would clearly create the equivalence class for any $$\displaystyle t$$ that generates $$\displaystyle z$$.

But what would you say about it's continuity? Is $$\displaystyle Arg(z)$$ 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 $$\displaystyle X/\sim$$ is compact (it's not too hard) and then you're homeomorphism is immediate.

• ejgmath

#### ejgmath

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