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Suppose we have a map $f:X \times Y \rightarrow Z$, where $X,Y$, and $Z$ are topological spaces. Are there any conditions on $X$,$Y$, and $Z$ that would allow one to determine that $F$ is continuous if it was known that it was continuous in each variable? It seems like there should be a theorem related to this.

By definition, a path homotopy $F: X \times I \rightarrow Y$ is continuous. What results in algebraic topology would not hold if we only required the map to be continuous in each variable? Would path homotopies not necessarily generate the fundamental group?

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    Even in the nice case $X=Y = \mathbb R$ the continuity in each variable is not sufficient for the continuity as a function of two variables - the same will hold if you take $X=Y=[0,1]$.2011-07-13
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    f(x,y)=$ \frac{2xy}{x^2+y^2}; f(0,0)=0 $is a standard counterexample; continuous for each of x,y separately , but not continuous (check the limit at $(0,0)$ , e.g., along the direction y=x ). You may have to use sequential continuity ( if you have first-countability), or just the "old-fashioned" way, by showing that the inverse image under f of an open set in Z is open in the product topology of $X \times Y \$2011-07-13

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