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Thread: Continuous functions and homeomorphisms

  1. #1
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    Continuous functions and homeomorphisms

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

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

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

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

    $\displaystyle g(a) = (g_1(a), g_2(a))$.

    Then g is continuous if and only if the functions

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

    are continuous.


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

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

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

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

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

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

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

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

    Note: This is my attemp to this problem. Any feedback or correction is welcomed.
    Last edited by aliceinwonderland; Jun 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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