Let $\gamma:[0,1]\to \Bbb C$ be a closed rectifiable curve and consider $\gamma^{-1}:[0,1]\to \Bbb C$ given by $\gamma^{-1}=\gamma(1-t)$. Show that trace($\gamma$)=trace($\gamma^{-1}$).
I tried to show that $\gamma$ and $\gamma^{-1}$ is equivalent, since equivalent paths has the same trace. However I found that they are not equivalent, since the $\gamma^{-1}(t)=\gamma\circ\phi$, where $\phi(t)=1-t$ doesn't satisfy the condition strictly increasing and also does not satisfy $\phi(0)=0,\phi(1)=1$.
Intuitively I believe this is true, but I am having trouble proving it. Could anyone kindly help? Thanks so much!
The trace of a curve $\gamma:[0,1]\to \Bbb C$ (as I remember it) is just the range: $$ \text{trace}(\gamma) = \{ \gamma(t) \mid t \in [0, 1] \} \, . $$ Then $\text{trace}(\gamma) = \text{trace}(\gamma^{-1})$ follows directly from $$ t \in [0, 1] \Longleftrightarrow 1- t \in [0, 1] \, . $$