# Path homotopy and separately continuous functions

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?

• The idea is that we want to describe a continuous one-parameter family of paths. A family of paths which happens to be a union of one-parameter families of points is something morally different and I shouldn't need to construct a counterexample to convince you of this (although I assume some pathological example exists). – Qiaochu Yuan Feb 1 '12 at 6:35

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 $F(1,t)$.

• I am having a hard time following your example. It looks like $F(s,0)=1$, $F(s,1)=-1$, $F(0,t)=\exp(-i\pi t)$, and $F(1,t)=\exp(-i\pi t)$. So where is the clockwise path from $1$ to $-1$? – user782220 Feb 1 '12 at 22:56
• @user782220: Sorry, the $s=1$ case was bogus. I think it should be correct now. – Nate Eldredge Feb 2 '12 at 1:43

One reason is that we want to be guaranteed that for any map $g: I \hookrightarrow I \times I$ with $g(0) = (0,0)$ and $g(1) = (1,1)$, the composition $F \circ g$ to be continuous. This guarantees that we can continuously deform portions of the path independently, and is one of the main properties used to prove that multiplication (path composition) in the fundamental group is associative.