different notions of continuous maps

**Sogan** · July 9th 2011, 03:48 AM

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

**girdav** · July 9th 2011, 04:03 AM

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.

**Sogan** · July 9th 2011, 04:16 AM

Originally Posted by girdav

$(f_x)^{-1}(U) = \left\{y\in Y, f(x,y)\in U\right\} = \left\{p_2(x,y), f(x,y)\in U\right\} =p_2(f^{-1}(U))$

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).

**girdav** · July 9th 2011, 04:30 AM

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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