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Questions about a f:X->Y being a bijective continuous function.

  1. Sep 9, 2012 #1
    1. The problem statement, all variables and given/known data
    For my proof, it tells me that f:X->Y is bijective.
    I understand that it is one-to-one and onto, but I just want to be clear about this from a neighborhood (open subset by our def) standpoint.
    Just to be clear: if f is bijective continuous, then that means the for all open neighborhoods "V" in Y, there is a unique open neighborhood "U" in X s.t. f-1(V) = U (which is open in X).

    Is that correct?

    I want to affirm the uniqueness part.
    Thank you!


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
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