That is correct. $$ \gamma(t) = -e^{-it} = e^{i(\pi - t)} \quad (0 \le t \le \pi) $$ is an arc on the unit circle with the argument of $\gamma(t)$ decreasing from $\pi$ to zero.
That is correct. $$ \gamma(t) = -e^{-it} = e^{i(\pi - t)} \quad (0 \le t \le \pi) $$ is an arc on the unit circle with the argument of $\gamma(t)$ decreasing from $\pi$ to zero.