Do the endpoints need to be fixed in order to define a homotopy?

Question

Basically, the title. Consider the function (homotopy?) $H\left(\theta, x\right)$ such that $H\colon\left[0,\pi/2\right]\times\left[-1,1\right]\to\mathbb{R}$ and

$$H\left(\theta, x\right)=f\left(x\right)+g\left(x\right)\cos\theta$$

Obviously, $H\left(0,x\right) = f\left(x\right)$ and $H\left(1,x\right) = f\left(x\right) + g\left(x\right)$. Now, we know that $g\left(-1\right) = 0$ so that one of the endpoints is fixed. However, $g\left(1\right)\neq 0$, in general, so that $H\left(0, 1\right)\neq H\left(1,1\right)$. Would this still classify as a homotopy? If not, is there another technical term I should be aware of?

Answer 1

Homotopies, in general, don't need to preserve endpoints. So your function $H$ would be a perfectly valid homotopy.

There is a class of homotopies which are required to preserve endpoints. We say paths $f, g \colon [0,1] \to X$ are "homotopic relative boundary/endpoints", sometimes denoted "$f \simeq g$ rel $\partial$" or "$f \stackrel{\partial}{\simeq} g$", if there exists a continuous function $H \colon [0, 1] \times [0, 1] \times X \to X$ such that

1. $H(0, x) = f(x)$ for all $x \in [0,1]$,
2. $H(1, x) = g(x)$ for all $x \in [0, 1]$,
3. $H(\theta, 0) = f(0) = g(0)$ for all $\theta \in [0,1]$,
4. $H(\theta, 1) = f(1) = g(1)$ for all $\theta \in [0,1]$.

Points 1 and 2 are all that's required for a general homotopy. Points 3 and 4 are what's additionally required for a "homotopy rel endpoints".