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Math Help - 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 _x : Y->A, f_x (y)=(x,y) and 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 f: X \times Y->A be cont. then \forall U \subset A we get an open set 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 f_x : Y->A and take some U \subset A.
    Now we have to proof, (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; July 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. f_x(y) =f(x,y), not (x,y) and we should have f^{-1}(U)\supset X'\times Y' (we know that f^{-1}(U) is open, but the open sets are not always product of open sets). If we denote by p_2 :X\times Y\rightarrow Y, p_2(x,y)=y, we have (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 p_2(X'\times Y') is.
    Last edited by girdav; July 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
    (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 y_0\in Y and V \subset A an open set which contains f_x(y_0)=f(x,y_0). By definition of the continuity and the product topology, we can find U_1\subset X and U_2\subset Y two open sets with (x,y_0)\in U_1\times U_2\subset f^{-1}(V). Now, for y\in U_2 we have (x,y)\in U_1\times U_2\subset f^{-1}(V) hence f(x,y)\in V.
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