Continuous injective curve has "direction of travel"

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 $$t_1 then it follows that $$s_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$$s_1 < s_2$$.

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 $$t_n \to t$$.
• I think you've skipped a little... how do you conclude $\pi$ is monotonic from the fact it's a bijection? If it's a continuous bijection this makes sense but then you need to justify continuity of $\hat \gamma ^{-1}.$ Commented Nov 25, 2020 at 2:49