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Continuous functions in topology

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data
    In topology, a f: X -> Y is continuous when

    U is open in Y implies that f^{-1}(U) is open in X

    Doesn't that mean that a continuous function must be surjective i.e. it must span all of Y since every point y in Y is in an open set and that open set must have a pre-image open in X, so there must be some x in X s.t. f(x) = y?

    What is wrong with my logic?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 28, 2007 #2

    Hurkyl

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    If you think it isn't true, the try and construct a counterexample. :smile:

    If it's true, this might help you see why. If it's false, analysis of the counterexample might help you spot the flaw.
     
  4. Oct 28, 2007 #3

    matt grime

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    So you're asking about the case when there is no x in X st... That would seem to be the empty set. Now what was the definition of topology?
     
  5. Oct 28, 2007 #4
    I think matt grime answered it, but my counter example is:

    f:R->R
    f(x) = abs(x)

    the set (-4, -2) is open but when when you pull it back you get the null set, which is open in the usual topology on R. I think I see now!
     
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