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All open maps homeomorphisms?

  1. Oct 5, 2013 #1
    I was told to prove that

    f: X -> Y is a homeomorphism iff it is an open map

    While I see that if f is a homeomorphism, it is certainly an open map, but is the implication in the other direction even true? Because I see no reason to believe it is.
     
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  3. Oct 5, 2013 #2

    WannabeNewton

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    That is false. I suspect they meant a bijective continuous map ##f## is a homeomorphism iff it is open iff it is closed.
     
  4. Oct 5, 2013 #3
    I think it should be "A bijective, continuous map ##f:X\to Y## is a homeomorphism iff it is an open map."

    For fun, prove "A bijective, continuous map ##f:X\to Y## is a homeomorphism iff it is a closed map." :tongue:

    Edit: WN got there first. Why you type so fast? :rofl:
     
  5. Oct 5, 2013 #4

    mathwonk

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    this false without some hypotheses. if X is an open subset of Y, the inclusion map is open but not a homeomorphism.

    if f is projection of the plane X onto the y - axis Y, this map is open but not a homeomorphism.

    If X and Y are the same set equipped with different topologies so that Y has more open sets than X, then the identity map is open but not a homeomorphism.

    etc.....
     
  6. Oct 5, 2013 #5

    lavinia

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    take a look at the open mapping theorem for complex analysis.

    Let X be any space and Y a point. Map X to Y.
     
  7. Oct 5, 2013 #6
    Oh, ok. But isn't that kind of trivial? If it is an open map then that is exactly the same (from what I can see) as saying that f-inverse is continuous. So basically that means I'm asked to show that f is a homeomorphism if f is continuous, f-inverse is continuous, and f is a bijection, but that's just what a homeomorphism is.
     
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