A doubt in Hatcher's explanation of reparametrization.

Question

Hatcher contains the following paragraph:

Define a reparametrization of a path $f$ to be composition $f\psi$ where $\psi:I\to I$ is any continuous map such that $\psi(0)=0$ and $\psi(1)=1$. Reparametrizing a path preserves its homotopy class since $f\psi\simeq f$ via the homotopy $f\psi_t$, where $$\psi_t(s)=(1-t)\psi(s)+ts$$ so that $\psi_0=\psi$ and $\psi_1(s)=s$.

What does it mean to say that reparametrization preserves the homotopy class of a path? Aren't $f$ and $f\psi$ the same curves anyway? Why would one mention $f\psi\simeq f$?

Answer 1

The images of $f$ and $f\circ\psi$ are the same, but they are two different maps $I \to X$. A path is a map $I \to X$, not just its image. So it needs to be checked that the maps $f$ and $f\circ\psi$ are homotopic.

Added Later: In particular, just because two paths have the same image, it does not mean they are homotopic. For example, the paths $I \to S^1$, $x \mapsto e^{2\pi ix}$ and $I \to S^1$, $x \mapsto e^{-2\pi i x}$ have the same image, but they are not homotopic.

Answer 2

Since a path in $X$ is defined to be a map $\gamma\colon I\to X$, then $f$ and $f\psi$ are different paths, since they are different maps. But even though these two paths are different, they have the same image.