Surjective Function

Jan 2010
32
1
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
 
Last edited:

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
\(\displaystyle f([t])=\exp^{2\pi ti}\) 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.
 
  • Like
Reactions: ejgmath

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
\(\displaystyle f([t])\) 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.
 
  • Like
Reactions: ejgmath
Jan 2010
32
1
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
Nov 2009
4,563
1,566
Berkeley, California
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.
 
  • Like
Reactions: ejgmath
Jan 2010
32
1
Yeh, I did consider that but the (revision) question is specifically looking for the whole continuous bijection business.