TITLE: Path homotopy and separately continuous functions
QUESTION [12 upvotes]: Two paths $f$ and $f'$ mapping the interval $I=[0,1]$ into $X$, are said to be path homotopic if they have the same initial point $x_0$ and the same final point $x_1$, and if there is a continuous map $F:I\times I\to X$ such that
$$F(s,0)=f(s)$$
$$F(s,1)=f'(s)$$
$$F(0,t)=x_0$$
$$F(1,t)=x_1$$
for each $s\in I$ and each $t\in I$.
My question is why is it required that $F$ be continuous on $I\times I$ and not separately continuous? The idea is that $F$ being only separately continuously means that for each fixed $t$ we have that $g(s)=F(s,t)$ is continuous in $s$. In other words $g(s)$ is a path. And for each fixed $s$ we have that $h(t)=F(s,t)$ is continuous in t. In other words $h(t)$ describes how the point $F(s,0)$ travels through time in a continuous path. These conditions or descriptions of $g(s)$ and $h(t)$ seem to capture the notion of "homotopic" to me. Is there some pathological reason why separately continuous is no good?

REPLY [14 votes]: This example should convince you that separately continuous is not the right definition for a homotopy.
Consider $S^1$ as the unit circle in the complex plane.   Take the map $F : I \times I \to X$ given by
$$F(s,t) = \begin{cases}
\exp(i \pi t), & 0 \le t \le s \le 1 \\
\underbrace{\exp \left(-i \pi \frac{(s+1)t - 2s}{1-s}\right)}_{= exp \left(i \pi \frac{(s+1)t - 2s}{s-1}\right)}, & 0 \le s < t \le 1 \\\\
\end{cases}$$
$F$ is separately continuous, $F(0,t) = \exp(-i \pi t)$, $F(1,t) = \exp(i \pi t)$, $F(s,0)=1$, $F(s,1)=-1$.  So this is a "homotopy" between a path from 1 to -1 which moves clockwise around the unit circle, and one which moves counterclockwise.  
In other words, with your definition, $S^1$ would be simply connected!  So using separate continuity would not give us a homotopy theory that fits our intuition.
Note that $F(s,t)$, as a function of $t$, is a path that moves counterclockwise along the unit circle at speed 1 for the first $s$ seconds, and then uses the remaining time to move clockwise all the way to -1.  As $s \to 1$, the clockwise trip is squeezed into a very short time, until at $s=1$ it disappears completely.  
The key point is that as $s \to 1$, we have $F(s,t) \to F(1,t)$ pointwise in $t$, but not uniformly in $t$, so the paths $F(s,t)$ and $F(1,t)$ are really not very similar looking.  For $s$ close to 1, you can't really justify calling $F(s,t)$ a "continuous deformation" of $F(1,t)$.