Results 1 to 6 of 6

Math Help - Surjective Function

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    32

    Surjective Function

    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
    Last edited by ejgmath; May 13th 2010 at 09:20 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by ejgmath View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2010
    Posts
    32
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    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 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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jan 2010
    Posts
    32
    Yeh, I did consider that but the (revision) question is specifically looking for the whole continuous bijection business.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Checking if a function is surjective
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: August 28th 2011, 07:57 AM
  2. surjective function
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: March 20th 2011, 04:07 AM
  3. surjective function
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: March 17th 2011, 07:45 AM
  4. Showing a function is not surjective
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: March 12th 2010, 04:16 PM
  5. surjective function
    Posted in the Pre-Calculus Forum
    Replies: 0
    Last Post: August 18th 2008, 07:26 AM

Search Tags


/mathhelpforum @mathhelpforum