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Math Help - Continuous functions and homeomorphisms

  1. #1
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    Continuous functions and homeomorphisms

    Let X and Y be topological spaces and f : X \longrightarrow Y a function. The graph of f is the set G_f = \{(x,f(x)) : x \in X\}.

    Prove that f is continuous if and only if the function \Phi : X \longrightarrow G_f given by \Phi(x) = (x,f(x)) is a homeomorphism where G_f has the topology relative to the product topology X \times Y.
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  2. #2
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    Quote Originally Posted by Inti View Post
    Let X and Y be topological spaces and f : X \longrightarrow Y a function. The graph of f is the set G_f = \{(x,f(x)) : x \in X\}.

    Prove that f is continuous if and only if the function \Phi : X \longrightarrow G_f given by \Phi(x) = (x,f(x)) is a homeomorphism where G_f has the topology relative to the product topology X \times Y.
    Definition. Let X and Y be a topological space; let h : X \rightarrow Y be a bijection.
    If both the function h and the inverse function h^{-1}:Y \rightarrow X are continuous, then h is called a homeomorphism.

    Lemma 1. Let g:A \rightarrow X \times Y be given by the equation

    g(a) = (g_1(a), g_2(a)).

    Then g is continuous if and only if the functions

    g_1:A \rightarrow X and g_2:A \rightarrow Y

    are continuous.


    ( \Rightarrow)
    Suppose f is a continuous function.
    Since f is a well-defined function, it is clear that \Phi : X \longrightarrow G_f defined by x \mapsto (x, f(x)) is a bijection. Let U \times V be an open set of G_f relative to the product topology  X \times Y containing an arbitrary point  (x_1, f(x_1)) \in G_f. It follows that

    \Phi^{-1}(U \times V) = U \cap f^{-1}(V).

    Since both an identity map and f are continuous functions, U and f^{-1}(V) are open, so is their intersection. It follows that \Phi^{-1}(U \times V) is open. Thus, \Phi is continuous.

    Now, we need to show that \Phi ^{-1} : G_f \longrightarrow X defined by (x, f(x)) \mapsto x is a continuous function. Let U be an open set containing an arbitrary point of x_1 \in X. Then, (\Phi ^{-1})^{-1}(U) = K, where K is an intersection of a neighborhood of (x_1, f(x_1)) \in U \times V \subset X \times Y with G_f. Specifically,

    K = (U \times M) \cap G_f = (U \times V) \cap G_f ,

    where M is a set f(U) \subset Y and V is an open set in Y containing M.

    Since G_f has the topology relative to the product topology X \times Y, K is open in G_f. Thus, \Phi ^{-1} is continuous.
    By the defintion given the above, \Phi is a homeomorphism.

    ( \Leftarrow)
    Suppose \Phi is a homeomorphism.
    Since an identity map is continuous and \Phi : X \longrightarrow G_f given by \Phi(x) = (x,f(x)) is continuous by hypothesis, f is continuous by lemma 1.

    Note: This is my attemp to this problem. Any feedback or correction is welcomed.
    Last edited by aliceinwonderland; June 15th 2009 at 08:25 AM. Reason: Plz check the modification for a set M.
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  3. #3
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    Cool, it seems to be ok to me.
    Thank you very much!!!
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