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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 $f\psi\simeq f$?

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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.

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Thanks a lot. But showing that $f$ and $f\psi$ are homotopic should involve proving the existence of a continuous mapping from $I\times I$ to $X$ with certain conditions. Are we really doing that here? How does showing that two paths are path homotopic in the domain say anything about the images of those paths? Is there even a continuous mapping to $X$? –  freebird Jul 13 at 17:15
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We are considering a map $I\times I \to X$, it is given by $(s, t) \mapsto (f\circ\psi_t)(s)$. –  Michael Albanese Jul 13 at 17:18
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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.

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