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Thread: Surjective Function

  1. #1
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    Surjective Function

    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 by ejgmath; May 13th 2010 at 08:20 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    $\displaystyle f([t])=\exp^{2\pi ti}$ 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.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by ejgmath View Post
    $\displaystyle f([t])$ 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 $\displaystyle \pi\circ\alpha\circ\xi:\mathbb{R}\to\mathbb{R}:t\m apsto \cos(2\pi[t])=\cos(2\pi t)$. 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.
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  4. #4
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    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?
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by ejgmath View Post
    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.
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  6. #6
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    Yeh, I did consider that but the (revision) question is specifically looking for the whole continuous bijection business.
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