In our class, we wrote the following:
Connected curves (1-dimensional manifolds) can be parametrised globally; $\overrightarrow{\gamma}: I \subseteq \mathbb{R} \to \Gamma \subseteq \mathbb{R}^3$, where $\overrightarrow{\gamma}$ is a $C^1$ map from interval $I$ to curve $\Gamma$. If $\dot{\overrightarrow{\gamma}} \neq \overrightarrow{0}$ on $I$ then we say $\overrightarrow{\gamma}$ is a regular parametrisation.
Curve $\Gamma$ has infinitely many regular parametrisations. Let $\overrightarrow{r}: [ a, b ] \to \Gamma$ and $\overrightarrow{\rho}: [\alpha, \beta] \to \Gamma$ be two regular parametrisations. Then there is a diffeomorphism $h = \overrightarrow{r}^{-1} \circ \overrightarrow{\rho}$ from $[\alpha, \beta]$ to $[a,b]$.
Because $\overrightarrow{\rho}' = \dot{\overrightarrow{r}}(h) \cdot h'$ and $\overrightarrow{\rho}' \neq \overrightarrow{0}$ on $[\alpha, \beta]$, it is also true that $h' \neq 0$ on $[ \alpha, \beta ]$, and $h'$ is therefore also a diffeomorphism.
If we know one parametrisation $\overrightarrow{r}$ for $\Gamma$, then all other parametrisations are of the form $\overrightarrow{r} \circ h$ for some diffeomorphism $h$.
My first question is the following: why is $h$, as defined in the second paragraph, necessarily a diffeomorphism?
Why are $h$ and $h^{-1}$ $C^1$? $\overrightarrow{r}$ and $\overrightarrow{\rho}$ are $C^1$ by definition, but their inverses are not necessarily; for example, if $\overrightarrow{r}$ is the standard parametrisation of a circle, then it's inverse is not even continuous.
Why is $h$ a bijection? I don't think even $\overrightarrow{r}$ is a bijection; we can parametrise a circle by moving around it multiple times and get a non-injective parametrisation this way.
My second question is why does the fact that $h' \neq 0$ on $[ \alpha, \beta ]$ immediately imply that $h'$ is also a diffeomorphism?