A further question on reparametrization. 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 $\psi_1(s)=s$. 

Here is what I don't understand about reprametrization:


*

*What does reparametrization achieve? Ultimately, the images of both $f$ and $f\circ \psi_t$ are the same for all $t\in[0,1]$

*I understand that $\psi$ can be continuously deformed into $1$, where $1$ is the identity mapping of $I\to I$. But the continuous deformation of $\psi$ to $1$ does not have any bearing on the continuous deformation of $(f\circ \psi)$ to $(f\circ 1)$, does it? To prove that $f\circ 1$ and $f\circ \psi$ are homotopic, a continuous mapping $(s,t)\to (1-t)(f\circ\psi)+t (f\circ 1)$ is defined anyway. So why should we focus on $\psi$ deforming continuously to $1$, if it has no bearing on homotopy whatsoever?
 A: To answer question 1, certainly a function is a different object than its image. The paths $f(s)=s$ and $g(s)=s^2$, $0 \le s \le 1$, are two different paths in $\mathbb{R}$, having the same image $[0,1]$. But, because they are different functions, they are therefore different paths. The parameterization can be thought of intuitively as a stopwatch ticking along to mark your progress as you walk along. For the $f(s)=s$ path, as the stopwatch ticks off from $0$ to $1$ you are moving at constant velocity. For the $g(s)=s^2$ path, you are starting out slowly and speeding up. Same image, different ways of moving on the image.
To answer question 2, your formula $(1-t)(f \circ \psi) + t (f \circ 1)$ is nonsense. You have not told us what the range of your function is, but presumably the range is some topological space $Y$ which need not have any kind of addition or scalar mutplication. So how do you propose to multiply a point of $Y$ by $1-t$ or by $t$? Instead, to get a homotopy from the path $s \mapsto f \circ \psi(s)$ to the path $s \mapsto f(s)$, you use exactly what is written, namely the homotopy $(s,t) \mapsto f \circ \psi_t(s)$.
