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Thread: different notions of continuous maps

  1. #1
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    different notions of continuous maps

    Hi,

    i just learned the notions of jointly and separately cont. functions.
    A function f: X x Y -> A is called
    1) jointly cont <=> f is cont. w.r.t. the product topology
    2) separ. cont <=> f$\displaystyle _x : Y->A, f_x (y)=(x,y)$ and $\displaystyle f_y$ are cont.

    So i tryed to show 1) => 2).

    But all my attempts failed.

    Can you help me?

    One of my attemts:
    Let $\displaystyle f: X \times Y->A$ be cont. then $\displaystyle \forall U \subset A$ we get an open set $\displaystyle X' \times Y':=f^{-1} (U) \subset X \times Y$ ., whereas X' and Y' are open in X, Y respectively.

    If we now consider the map $\displaystyle f_x : Y->A $ and take some U \subset A.
    Now we have to proof, $\displaystyle (f_x) ^{-1}(U)=\{y \in Y : (x,y) \in A\}$ is open in Y. But why is this correct in all top. spaces?

    Regards
    Last edited by Sogan; Jul 9th 2011 at 04:03 AM.
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  2. #2
    Super Member girdav's Avatar
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    Re: different notions of continuous maps

    Yes, it's indeed the idea.
    There are some typos in your post. $\displaystyle f_x(y) =f(x,y)$, not $\displaystyle (x,y)$ and we should have $\displaystyle f^{-1}(U)\supset X'\times Y'$ (we know that $\displaystyle f^{-1}(U)$ is open, but the open sets are not always product of open sets). If we denote by $\displaystyle p_2 :X\times Y\rightarrow Y, p_2(x,y)=y$, we have $\displaystyle (f_x)^{-1}(U) = \left\{y\in Y, f(x,y)\in U\right\} = \left\{p_2(x,y), f(x,y)\in U\right\} \supset p_2(f^{-1}(U))$ and we are done since we know what $\displaystyle p_2(X'\times Y')$ is.
    Last edited by girdav; Jul 9th 2011 at 04:24 AM.
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  3. #3
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    Re: different notions of continuous maps

    Quote Originally Posted by girdav View Post
    $\displaystyle (f_x)^{-1}(U) = \left\{y\in Y, f(x,y)\in U\right\} = \left\{p_2(x,y), f(x,y)\in U\right\} =p_2(f^{-1}(U))$
    Hello,

    I have two questions about your Proof. First, the last equation isn't so obvious for me. Since on the LHS we have a fixed 'x' (depending on the choice of f_x). On the RHS we consider all 'x' (as preimage of f).
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  4. #4
    Super Member girdav's Avatar
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    Re: different notions of continuous maps

    It's wasn't obvious for you because it was false.
    To show the result, let $\displaystyle y_0\in Y$ and $\displaystyle V \subset A$ an open set which contains $\displaystyle f_x(y_0)=f(x,y_0)$. By definition of the continuity and the product topology, we can find $\displaystyle U_1\subset X$ and $\displaystyle U_2\subset Y$ two open sets with $\displaystyle (x,y_0)\in U_1\times U_2\subset f^{-1}(V)$. Now, for $\displaystyle y\in U_2$ we have $\displaystyle (x,y)\in U_1\times U_2\subset f^{-1}(V)$ hence $\displaystyle f(x,y)\in V$.
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