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Math Help - locally compact and proper functions

  1. #1
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    locally compact and proper functions

    Hello,

    i try to solve following exercise:

    Let f:X-> Y be an continuous immersion. X a Hausdorff-space and Y locally compact, i.e. Y is a Hausdorff-space with the property that every point y \in Y has a nbh. U with closure(U) compact.
    Show that f is proper iff f^(-1) (K) \subset X is compact for all K \subset Y compact.


    "=>" This direction was not so difficult. I could solve it.

    But the other direction "<=" i couldn't solve.

    We have shown before, that "f is proper" is equivalent to "f is closed" and to "f is an embedding".

    I try to show that f is closed, since i think this must be the most easy way.

    Have you some help for me.

    Thanks a lot
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  2. #2
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    I have a solution, but i'm not sure whether it is correct:

    "<="
    We want to show f is proper <=> f is closed.

    Let V be a closed subset of X.

    f(V) is closed in Y <=> Y-f(V) is open in Y.

    so let y \in Y-f(V) a arbitrary point. We know Y is Hausdorff, hence for every point v \in f(V) we have a open set U_v of v and a open set U of y, s.t.
     U_v \bigcap U= \emptyset

    (*)hence it exist a nbh U of y , s.t. U \bigcap f(V) = \emptyset

    hence f(V) is closed.


    I'm not sure whether this prove is correct.
    I mean, i havnÄt use locally compactness of Y. So i think this can't be correct.
    the part with (*) above isn't correct, or? how can i involve locally compactness in that point, such that my proof is correct?

    Regards
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Sogan View Post
    I have a solution, but i'm not sure whether it is correct:

    "<="
    We want to show f is proper <=> f is closed.

    Let V be a closed subset of X.

    f(V) is closed in Y <=> Y-f(V) is open in Y.

    so let y \in Y-f(V) a arbitrary point. We know Y is Hausdorff, hence for every point v \in f(V) we have a open set U_v of v and a open set U of y, s.t.
     U_v \bigcap U= \emptyset

    (*)hence it exist a nbh U of y , s.t. U \bigcap f(V) = \emptyset

    hence f(V) is closed.


    I'm not sure whether this prove is correct.
    I mean, i havnÄt use locally compactness of Y. So i think this can't be correct.
    the part with (*) above isn't correct, or? how can i involve locally compactness in that point, such that my proof is correct?

    Regards
    Yeah, it's not so clear to me why you can do what you said you can. But how about this:

    I'm a little unsure what you're question actually is. Is it that f is proper implies that f is closed?
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  4. #4
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    Hello,

    thanks for your help.

    No thants not my question. I want to show following:

    if f^(-1)(K) is compact for every compact K => f is proper.

    This statement is what i want to prove
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Sogan View Post
    Hello,

    thanks for your help.

    No thants not my question. I want to show following:

    if f^(-1)(K) is compact for every compact K => f is proper.

    This statement is what i want to prove
    Isn't that the definition of a proper map?
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  6. #6
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    Oh ok. we have a different definition of proper map:

    f is proper iff f x id_z : X x Z->Y x Z is closed.

    f in C(X,Y)

    but both definitions are equivalent. That is what we should show in this exercise.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Sogan View Post
    Oh ok. we have a different definition of proper map:

    f is proper iff f x id_z : X x Z->Y x Z is closed.

    f in C(X,Y)

    but both definitions are equivalent. That is what we should show in this exercise.
    What is Z? Any space?
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  8. #8
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    yes any topological space
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Sogan View Post
    yes any topological space
    Ok, one last question. Are we assuming that X,Y are Hausdorff and Y locally compact?
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  10. #10
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    Yes you are right. X,Y are Hausdorff and Y is locally compact
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