I'm having some trouble with the following exercise:
Let $\alpha:[a,b]\to \mathbb R^n$ and $\beta:[c,d]\to \mathbb R^n$ be two regular, injective curves with the same trace in $\mathbb R^n$.
Prove that $\beta^{-1}\circ\alpha:[a,b] \to [c,d]$ is differentiable and it's derivative is never zero.
I tried to prove that, for any $k\in [a,b]$, the limit: $$\lim_{t\to k} \frac{\beta^{-1}(\alpha(k)) - \beta^{-1}(\alpha(t))}{k-t}$$
exists. I wasn't able to conclude anything because we don't know if $\beta^{-1}$ is continuous or differentiable.
How can this be done?
Let $J = \alpha([a,b]) = \beta([c,d]) \subset \mathbb R^n$ be the common trace of both curves. Then $\beta : [c,d] \to J$ is a continuous bijection from a compact to a Hausdorff space, thus a homeomorphism. Hence $\beta^{-1 } : J \to [a,b]$ is a homeomorphism. Similarly $\alpha :[a,b] \to J$ is a homeomorphism. This shows that $$u = \beta^{-1} \circ \alpha : [a,b] \to [c,d]$$ also is a homeomorphism.
For $t \ne t_0$ we have $$\frac{\alpha(t) - \alpha(t_0)}{t - t_0} = \frac{\beta(u(t)) - \beta(u(t_0))}{t - t_0} = \frac{\beta(u(t)) - \beta(u(t_0))}{u(t) - u(t_0)} \cdot \frac{u(t) - u(t_0)}{t - t_0} . \tag{1} $$ Note that $u(t) - u(t_0) \ne 0$ because $u$ is injective and $\beta(u(t)) - \beta(u(t_0)) = \alpha(t) - \alpha(t_0) \ne 0$ because $\alpha$ is injective. Taking the norm gives us $$\left\lVert \frac{\alpha(t) - \alpha(t_0)}{t - t_0} \right\rVert = \left\lVert \frac{\beta(u(t)) - \beta(u(t_0))}{u(t) - u(t_0)} \right\rVert \cdot \left\lvert \frac{u(t) - u(t_0)}{t - t_0} \right\rvert. \tag{2} $$
Thus $$\left\lvert \frac{u(t) - u(t_0)}{t - t_0} \right\rvert = \frac{\left\lVert \frac{\alpha(t) - \alpha(t_0)}{t - t_0}\right\rVert}{\left\lVert \frac{\beta(u(t)) - \beta(u(t_0))}{u(t) - u(t_0)} \right\rVert}. \tag{3} $$ Since $u(t) \to u(t_0)$ as $t \to t_0$, we conclude that $$\lim_{t \to t_0} \left\lvert \frac{u(t) - u(t_0)}{t - t_0} \right\rvert = q := \frac{\left\lVert \alpha'(t_0)\right\rVert}{\left\lVert \beta'(u(t_0))\right\rVert} \ne 0 .\tag{4}$$
$u$ may be strictly increasing (in which case $u(a) = c$ and $u(b) = d$) or strictly decreasing (in which case $u(a) = d$ and $u(b) = c$).
In the strictly increasing case we have $$\left\lvert \frac{u(t) - u(t_0)}{t - t_0} \right\rvert = \frac{u(t) - u(t_0)}{t - t_0} \tag{5}$$ which implies $$\lim_{t \to t_0} \frac{u(t) - u(t_0)}{t - t_0} = q \ne 0 .\tag{6}$$ In the strictly decreasing case we have $$\left\lvert \frac{u(t) - u(t_0)}{t - t_0} \right\rvert = -\frac{u(t) - u(t_0)}{t - t_0} \tag{7}$$ which implies $$\lim_{t \to t_0} \frac{u(t) - u(t_0)}{t - t_0} = -q \ne 0 .\tag{8}$$