If $f: I \to X$ is a path and $c: I \to X$ the constant path $c(s)=f(1)$, how is $f \cdot c$ a reparametrization of $f$

Question

From Algebraic Topology by Hatcher:

Definition of reparametrization:

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Paragraph in question:

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If a reparametrization is defined to be a composition $f \varphi$ where $\varphi: I \to I$ satisfies $\varphi(0)=0$ and $\varphi(1)=1$, how is $f\cdot c$ a reparametrization if $c$ does not satisfy the definition?

Answer 1

$f\cdot c(t)= \begin{cases} f(2t) & 0\le t\le 1/2 \\ c(2t-1) & 1/2\le t\le 1 \end{cases}$

so in fact

$f\cdot c(t)= \begin{cases} f(2t) & 0\le t\le 1/2 \\ f(1) & 1/2\le t\le 1 \end{cases}$

And since

$\phi(t)= \begin{cases} 2t & 0\le t\le 1/2 \\ 1 & 1/2\le t\le 1 \end{cases}$,

a simple calculation gives $\phi\circ f=c\cdot f,\ $ as desired.

The other case is done in exactly the same way.