Continuous injective curve has "direction of travel"

Question

Let $\gamma,\hat{\gamma}:[0,1]\to\mathbb{R}^n$ be continous, injective curves such that $\gamma([0,1])=\hat{\gamma}([0,1])$. How can I show that if the two curves start in the same point and we have that $\gamma(t_1)=\hat{\gamma}(s_1)$, $\gamma(t_2)=\hat{\gamma}(s_2)$ and $t_1<t_2$ then it follows that $s_1<s_2$.

Answer 1

Since $\gamma, \hat{\gamma}$ are bijections onto their (same) ranges, we see that $\pi = \hat{\gamma}^{-1} \circ \gamma$ is a bijection $\pi:[0,1]\to[0,1]$. In particular it must be monotonic, and since $\gamma(0)= \hat{\gamma}(0)$ it must be increasing.

Hence given the conditions in the question we see that $t_1 < t_2$ iff$s_1 < s_2$.

Addendum:

As @AnthonyCarapetis pointed out, I skipped over a detail.

I need to show that $\hat{\gamma}^{-1}$ is indeed continuous.

Suppose $\hat{\gamma}(t_n) \to \hat{\gamma}(t)$ but $t_n \not\to t$ ($t_n,t$ are well defined since $\hat{\gamma}$ is injective). Then there is some $\delta >0$ such that $|t_n-t| \ge \delta$ infinitely often and hence $t_n \overset{K}{\to} t^* \neq t$ for some subsequence $K$. Then $\gamma(t_n) \overset{K}{\to} \gamma(t^*) \neq \gamma(t)$ (again because of injectivity) which is a contradiction and so $t_n \to t$.