Let $X$ be some topological space, let $a,b,c$ be points in $X$, let $f$ be the path from $a$ to $b$, and let $g$ be the path from $b$ to $c$. Define $h = f * g$. Then $\overline{h} = \overline{g} * \overline{f}$
By definition $\overline{h} = h(1-t)$, so I just need to verify $h(1-t) = g(1-t) * f(1-t)$ by computing the LHS and RHS and then comparing. Let's do the RHS first:
$$g(1-t) * f(1-t) = \begin{cases} g(2(1-t)), & t \in [0,\frac{1}{2}] \\ f(2(1-t)-1), & t \in [\frac{1}{2},1] \end{cases}$$
Simple enough. However, I run into a little trouble when computing the LHS, specifically a "domain" issue:
$$h(1-t) = \begin{cases} f(2(1-t)), & t \in [0,\frac{1}{2}] \\ g(2(1-t)-1), & t \in [\frac{1}{2},1] \end{cases}$$
According to the above, when $t=1$ we get $h(1) = g(2(1-1)-1) = g(-1)...$ What did I do wrong? $1-t$ should have a different domain, right? I thought $[-1,0]$ might work, but it didn't.
PS
When reading through my post again, I just realized that I also have domain issues associated with the RHS.
Let's start at the beginning:
$$h(t) = (f \ast g)(t) := \begin{cases} f(2t) & t \in [0,\frac12] \\ g(2t-1) & t \in [\frac12,1] \end{cases}$$
So
$$\bar{h}(t) = h(1-t) = \begin{cases} f(2(1-t)) = f(2-2t) & 1-t \in [0,\frac12] \\ g(2(1-t)-1)= g(1-2t) & 1-t \in [\frac12,1] \end{cases}$$
which after rewriting the conditions ($1-t \in [0,\frac12]$ iff $t \in [\frac12,1]$ etc.) becomes:
$$\bar{h}(t) = \begin{cases} g(1-2t) & t \in [0,\frac12] \\ f(2-2t) & t \in [\frac12,1] \end{cases}$$
Now $\bar{g}(t) = g(1-t)$ etc. so
$$(\bar{g} \ast \bar{f})(t) = \begin{cases} \bar{g}(2t) = g(1-2t) & t \in [0,\frac12]\\ \bar{f}(2t-1) = f(1-(2t-1)) =f(2-2t) & t \in [\frac12,1] \end{cases} $$
Which checks out with $\bar{h}$.