Let $X$ be a topological space with $\alpha,\beta:[0,1]\rightarrow X$ two paths such that $\alpha(0)=x_0,~\alpha(1)=\beta(0)=x_1 $ and $\beta(1)=x_2$. Suppose I would like to evaluate the expression $\alpha^{-1}*\beta^{-1}$. Should I first compute the inverses and then multiply the paths or multiply the paths and take their inverses ? Here are the results of performing the calculation in both routes.
Invert then multiply: \begin{align}\alpha^{-1}*\beta^{-1} &=\alpha(1-s)*\beta(1-s) \\ &=\begin{cases}\alpha(2-2s)&&0\leq s\leq 1/2\\\beta(1-2s)&&1/2 \leq s \leq 1\end{cases}\end{align}
Multiply then invert: $$\begin{align}\alpha^{-1}*\beta^{-1}&= \begin{cases}\alpha^{-1}(2s)&&0 \leq s \leq 1 \\ \beta^{-1}(2s-1) && 1/2 \leq s \leq 1\end{cases} \\&= \begin{cases}\beta(1-2s)&&0 \leq s \leq 1/2 \\ \alpha(2-2s)&& 1/2 \leq s \leq 1\end{cases}\end{align}$$
Just incase, my definitions (Nakahara, 2003) of inverting and multiplying are $$\alpha*\beta=\begin{cases}\alpha(2s)&&0\leq s \leq 1/2\\\beta(2s-1)&& 1/2 \leq s \leq 1\end{cases}$$ and $\alpha^{-1}(s)=\alpha(1-s)$. I would think that first you multiply and then you invert, since this would give $(\alpha*\beta)^{-1}=\alpha^{-1}*\beta^{-1}$ which makes sense.
You invert and then multiply.
The other way around is written $(\alpha * \beta)^{-1}$ and you can check that $(\alpha * \beta)^{-1} = \beta^{-1} * \alpha^{-1}$