Functions continuous in each variable 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?
 A: As is the case with two others who have responded (at the time I wrote this), I don't have an answer to your specific questions (conditions on $X$, $Y$, and $Z$; homotopy analogs for separately continuous maps). However, Piotrowski's 1996 survey paper or my 2005 sci.math post (which contains some references not given in Piotrowski's paper -- [2], [4], and [6]) might have something of interest to you or lead you to a relevant reference.
Zbigniew Piotrowski, "The genesis of separate versus joint continuity", Tatra Mountains Mathematical Publications 8 (1996), 113-126. [MR 98j:01026; Zbl 914.01007]
http://people.ysu.edu/~zpiotrowski/papers/genesisseperatevsjoint.pdf
sci.math -- "Continuity in each variable vs. joint continuity" (4 June 2005)
http://groups.google.com/group/sci.math/msg/a1b3752adec7650e
A: Letting $S^1=\{z\in\mathbb{C}\colon\vert z\vert=1\}$ be the unit circle, consider the map $F\colon S^1\times I\to S^1$ given by
$$
F(e^{2\pi\theta i},s) = e^{2\pi\theta^si}
$$
for $0 < \theta\le 1$ and $s\in I$. Then $F$ is continuous in each variable, $F(z,1)=z$ and $F(z,0)=1$. So, if you only required continuity in the individual variables, the circle would be contractible. More generally, every topological space would have trivial fundamental group. Suppose that $X$ is a topological space and $\gamma\colon I\to X$ is a closed curve. Define $F\colon I\times I\to X$ by $F(x,s)=\gamma(x^s)$ for $x,s\in I$ and $s > 0$, and $F(x,0)=\gamma(0)=\gamma(1)$. Then $\gamma$ is null-homotopic (relative to ${0,1}$). So, the fundamental group collapses to the trivial group, as do all the higher homotopy groups.
