This exercise come frome the Hatcher's book, in section 1.1. Given a map $f:X\rightarrow Y,$ and a path $h:I\rightarrow X$ from $x_0$ to $x_1$, show that $f_*\beta_h=\beta_{fh}f_*$.
My attempt. Let $[g]\in \pi_1(X,x_0)$, then $f_*\beta_h[g]=f_*[hg\bar h]=[f\circ(hg\bar h)]$, $\beta_{fh}f_*[g]=\beta_{fh}[fg]=[(fh)(fg)(\overline{fh})]$. How can I get this two are equal.
If $\alpha,\beta:I\to X$ are two paths with $\alpha(1)=\beta(0)$ then the path composition is defined as:
$$\alpha\beta(t):=\begin{cases} \alpha(2t) &\text{if } 0\le t\le 1/2 \\ \beta(2t-1) &\text{if } 1/2\le t\le 1 \end{cases}$$
So if $f:X\to Y$ is a function then
$$f\circ(\alpha\beta)(t)=\begin{cases} f\circ\alpha(2t) &\text{if } 0\le t\le 1/2 \\ f\circ\beta(2t-1) &\text{if } 1/2\le t\le 1 \end{cases}$$
and
$$(f\circ\alpha)(f\circ\beta)(t)=\begin{cases} f\circ\alpha(2t) &\text{if } 0\le t\le 1/2 \\ f\circ\beta(2t-1) &\text{if } 1/2\le t\le 1 \end{cases}$$
Meaning $f\circ(\alpha\beta)$ is literally equal to $(f\circ\alpha)(f\circ\beta)$. By induction this extends to any number of (compatible) paths, regardless of the order (bracketing) of composition.