1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sin t = 0 homeomorphism

  1. Sep 23, 2006 #1
    Show that tan: (-pie/2,pie/2)->R is a homeomorphism where tan = sin/cos

    To show that f and f^-1 are cts, it seems trivial from a sketch but how do you do it?

    For 1-1 tan(x) = tan(y)

    Need to knwo x =y

    tan(x) = sinx.cosx = siny/cosy = tany

    => sixcosy = sinycosx

    this gets you sin(x-y) = 0
    But x-y = pie

    What's wrong here?

    Onto is obvious
  2. jcsd
  3. Sep 23, 2006 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    There are infinitely many solutions to the equation sin t = 0... but not many that are consistent with the conditions of your problem.

    (Incidentally, it's not just a homeomorphism, but also a diffeomorphism. That's easier to prove, since we're in dimension 1. :wink:)
  4. Sep 23, 2006 #3
    Then it must be 0 because the domain is the interval (-pie/2,pie/2)

    So how do I show that f and f^-1 are cts?
  5. Sep 24, 2006 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    How have you defined them? e.g. showing tan to be continuous is trivial if you've defined sine and cosine via power series (or are allowed to use their power series)
  6. Oct 9, 2006 #5
    So for onto
    for all y in R there exists an x in (-pie/2,pie/2) s.t. f(x)=y

    WTS: tanx = y

    Let y = siny/cosy = tany

    let x = y => y = tany = tanx

    is that sufficient? y could be inf.
  7. Oct 9, 2006 #6

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    How can y be 'inf'? Neither is infinity in R, nor is pi/2 (pi does not have an e) in the interval (-pi/2,pi/2).
  8. Oct 9, 2006 #7
    So then how would you show f^-1 is cts ?
  9. Oct 16, 2006 #8
    Okay let me try again.

    To show that f is cts pick an open subset G in R. then f-1 = arc tanx: R -> -pie/2, pie/2)

    Suppose G is the interval (a,b).

    Let f-1(G) = A. Thus A is contained in (-pie/2,pie/2)

    f-1(G) is an element of (f-1(a),f-1(b))=A

    which is an open subset and therefore cts.

    To show that the inverse of f is cts.

    we want to show that for any open G in (-pie/2,pie/2) thus f-1(G) = A where A is open.

    IF G = (a,b) then f-1(G) is an element of (f-1(a),f-1(b)) = (tana,tanb) which is an open interval.

    and hence cts.

    Is this correct?
  10. Oct 17, 2006 #9
    Okay, is there anything wrong with it?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?