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Let $X$, $Y$, $Z$ be arbitrary topological spaces and $f:X\times Y \rightarrow Z$ be an arbitrary map. Then, is it true that $f$ is continuous iff for every $y \in Y$ $f(\cdot,y):X\rightarrow Z$ is continuous and for every $x \in X$ $f(x, \cdot):Y\rightarrow Z$ is continuous? If not, on what occasion does it not hold? Does ($\Rightarrow$) or ($\Leftarrow$) hold?

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up vote 2 down vote accepted

It's true that if $f$ is continuous then the maps $g_y\colon X\to Z$, $g_y(x)=f(x,y)$ and $h_x\colon X\to Z$, $h_x(y)=f(x,y)$ are continuous. Indeed, fix $y\in Y$ and $O$ an open subset of $Z$. Then for all $x\in X$ we can find a neighborhood $V$ of $(x,y)$ in $X\times Y$ such that $f(V)\subset O$. By definition of the product topology, we can assume that $V=O_1\times O_2$, where $O_1$ is an open subset of $X$ and $O_2$ an open subset of $Y$. Then $g_y(O_1)\subset f(O_1\times O_2)\subset O$.

For the converse, look at the case $X=Y=Z=\mathbb R$ with the usual topology, and $$f(x,y)=\begin{cases}\frac{xy}{x^2+y^2}&\mbox{ if }(x,y)\neq (0,0)\\\ 0&\mbox{ if }(x,y)=(0,0)\end{cases}.$$

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Could you show me (the sketch of) the proof of ($\Rightarrow$) ? – Pteromys Feb 7 '12 at 9:04

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