When studying some basic differential geometry, I had to prove the following statement:
If two curves $\alpha: I_\alpha \rightarrow \mathbb{R}^n$ and $\beta :I_\beta \rightarrow \mathbb{R}^n$ determine the same route ($\alpha \sim \beta$ ) and $\alpha$ is regular, then $\beta$ is regular.
However, the definitions and conventions my course notes uses are somewhat different than what I find online. We say that two differentiable curves $\alpha: I_\alpha \rightarrow \mathbb{R}^n$ and $\beta :I_\beta \rightarrow \mathbb{R}^n$ determine the same route ($\alpha \sim \beta$) if there exists a differentiable bijection $f:I_\alpha \rightarrow I_\beta $ such that $\forall t \in I_\alpha:f'(t) > 0$, and such that $\alpha = \beta \ \circ \ f$. Then $\beta$ is also called a reparametrization of $\alpha$.
Also, by regular we mean a curve $\alpha$ that is of class $C^1$ (so differentiable and $\alpha'$ continuous ) and $\forall t \in I_\alpha: \alpha'(t) \neq 0$.
I have already shown that $\sim$ is an equivalence relation, and the only non trivial part there was the symmetry property.To prove that property I needed two conditions on the differentiable bijection $f$:
- $\forall t \in I_\alpha: f'(t) \neq 0$
- $f^{-1}$ is continuous
The first condition is conveniently given by the definition of $\sim$. However, a continuous bijection does not need to have a continuous inverse. But despite that fact, in this case, it will be true that $f^{-1}$ is continuous because $I_\alpha$ is given to be an open interval (this I have shown by using the intermediate value theorem).
So because these two conditions are fullfilled I know that $(f^{-1})'(t) = 1/f'(f^{-1}(t))$.
Now, in order to prove the above statement, I began by using the chain rule:
\begin{align} \alpha = \beta \circ f &\Rightarrow \beta = \alpha \circ f^{-1} \\ & \Rightarrow \forall t \in I_\beta: \beta'(t) = \frac{\alpha'(f^{-1}(t))}{f'(f^{-1}(t))} \end{align} Now since $\alpha$ is regular, I know that $\alpha'$ is continuous. If I can show that $f'$ is also continuous, I am done. I know that the derivative of a function will not be continuous in general, but my intuition says that in this case it will be true. It will probably have something to do with the fact that $f$ is a bijection on an open interval, just like the same reasoning I used for the inverse. However, I could not find an analogous reasoning for this exercise.
One last thing: I find it quite odd that for the proof that $\sim$ is an equivalence relation I had to come up with a rather non trivial way of showing that $f^{-1}$ is continuous. It feels like I am missing some crucial elements of the theory, with the result being that my proofs are rather long. These exercises were meant as an introduction to differential geometry, so I don't think they are meant to be this long.
It is a good question whether a differentiable bijection $f: I_\alpha\to I_\beta$ with $f'(x)>0$, where $I_\alpha$ and $I_\beta$ are open intervals in ${\mathbb R}$, would automatically imply that $f'$ is continuous.
I back-engineered this and found the following counter-example. $$ f(x)= \begin{cases} 3x+x^2\sin\frac{1}{x} & x\in (-1, 0)\cup (0, 1),\\ 0 & x=0. \end{cases} $$
We can directly compute that $f'(0) = 3$ from the definition of derivative. Also for $x\neq 0$, $f'(x)=3+2x\sin\frac{1}{x}-\cos\frac{1}{x}>3-2-1=0$ and we have a strict inequality since $|x|<1$. Therefore $f$ is differentiable, strictly increasing, and maps $(-1,1)$ bijectively to its image $(f(-1), f(1))$.
But $f'$ is not a continuous function, because of the $-\cos\frac{1}{x}$ term.
(Yes, it is still true that $f^{-1}$ is continuous and differentiable.)