example of homotopy which is not path homotopy

Question

Can someone give me a simple, concrete example of a homotopy, which is not a path homotopy?

Let $f, f'$ be continuous maps from $X$ to $Y$, and let $F: X \times I\to Y$ a continuous map such that $F(x,0) = f(x)$ and $F(x, 1) = f'(x)$. Then $f$ is homotopic to $f'$.

I understand path homotopies (the path can be deformed but the endpoints stayed fixed), but now I don't see what a homotopy is. I think the notation is what is messing me up, so can you explain what are the spaces $X$, $Y$ in your example? And what are your maps $f$, $f'$?

Answer 1

To match your notation, let $X=\mathbb{R}$ and let $Y=\mathbb{R}$ as well. Then we can define two maps $f:\mathbb{R} \to \mathbb{R}$ by $f(x)=0$ and $f':\mathbb{R} \to \mathbb{R}$ by $f'(x)=x$. The homotopy between the maps is given by $F: \mathbb{R} \times I \to \mathbb{R}$ by $F(x,t)=tx$.

We can see that $F(x,0)=0\cdot x = 0 = f(x)$ and $F(x,1) = 1 \cdot x = x = f'(x)$.

This homotopy "shrinks" the identity map down to the zero map.

A homotopy is more general than a path homotopy. A homotopy continuously deforms one function to another function, whereas a path homotopy continuously deforms path functions.

Answer 2

Homotopy between paths is a particular case of homotopy between maps. Paths in $X$ are maps from $I$ into $X$. So a homotopy between two paths $$\sigma,\gamma:I\to X$$ is a continuous map $H:I\times I\to X$ such that $H(s,0)=\sigma(s)$ and $H(s,1)=\gamma(s)$, for all $s\in I$.

But if the paths have the same end points, that is, $\sigma(0)=\gamma(0)$ and $\sigma(1)=\gamma(1)$, then you also require that for all $t\in I$, $H(0,t)=\sigma(0)$ and $H(1,t)=\sigma(1)$. This is called a based homotopy.