Let $f$ be a path and $\overline{f}=f(1-t)$. Show $f$~$g$ iff $\overline {f}$~$\overline{g}$ (by ~, I mean homotopic with respect to {$0,1$}.
My attempt: ($\Rightarrow$) Since $f$~$g$, there is a homotopy $F$ such that $F(s,0) = f(s)$ and $F(s,1) = g(s)$. Define $\overline {F} = F(1-s, t)$. Then clearly $\overline{F}$ satisfies the appropriate continuity requirements for a homotopy and $\overline{F}(s,0) = F(1-s,0) = f(1-s)= \overline{f}(s)$, and we have $\overline{F}(s,1) = F(1-s,1) = g(1-s)= \overline{g}(s)$. Now we just check that $\overline{F}(0,t)$ is the same $\forall t$ and that $\overline{F}(1,t)$ is the same $\forall t$. But $\overline{F}(0,t) = F(1,t)$, which is the same $\forall t$ by hypothesis, and $\overline{F}(1,t) = F(0,t)$, which is the same $\forall t$ by hypothesis. The ($\Leftarrow$) direction is similar.
Does this proof look right?
Looks good to me. Once you get the correct homotopy for the inverse paths (which you did), then the rest does follow quickly.
One nit-pick, if you used $t$ for your original parameter I would use $s$ for your homotopy parameter. It was a little confusing until I realized that $t$ had now become your homotopy parameter once you began the proof.