This is a question from Pressley's Elementary Differential Geometry Second Edition:
If $\tilde{\gamma}$ is a reparametrization of $\gamma$, and $\hat{\gamma}$ is a reparametrization of $\tilde{\gamma}$, then $\hat{\gamma}$ is a reparametrization of $\gamma$.
The solution given is:
If $\tilde{\gamma}(t) = \gamma(\phi(t))$ and $\hat{\gamma}(t) = \tilde{\gamma}(\psi(t))$, where $\phi$ and $\psi$ are reparametrization maps, then $\hat{\gamma}(t) = \gamma((\phi \circ \psi)(t))$ and $\phi \circ \psi$ is a reparametrization map because it is smooth and $\frac{d}{dt}(\phi(\psi(t)) = \dot{\phi}(\psi(t))\dot{\psi}(t) \not = 0$ as $\dot{\phi}$ and $\dot{\psi}$ are both $\not = 0$.
This book defined a reparametrization by its reparametrization map, which is a smooth, bijective function whose inverse is also smooth. Clearly, the composition of two smooth bijective functions, $\phi$ and $\psi$ have to be smooth and bijective. But why does it have to have non-zero derivatives? The definition of smooth only required the derivative to exist.
From what I can tell, the last step is intended to check if the reparametrization is regular, but since we are not given that $\gamma$ is regular, why is it necessary to check whether the repatametrization is regular?
Moreover, why is the reparametrization $\hat{\gamma}(t) = \gamma((\phi \circ \psi)(t))$? My understanding of this question is that I need to first transform the parameters to that of $\tilde{\gamma}$, then to $\hat{\gamma}$, so instead, I would have $\hat{\gamma}(t) = \gamma((\psi \circ \phi)(t))$ as my reparametrization. I've definitely gotten the concept wrong.
This is based on @TedShifrin's comment:
Deriving the equality $t = f^{-1}(f(t))$ gave $1=\frac{d}{dt}f^{-1}(f(t))\frac{d}{dt}f(t)$.
Then if $\frac{d}{dt}f(t) = 0$ at some point $t$, for some reparametrization map $f$, and since $\forall r, 1 \not = r \cdot 0 = 0 $, we have that $\frac{d}{dt}f^{-1}(f(t))$ cannot exist. This contradicts the fact that the inverse of a reparametrization map need to be smooth, so we have that $\frac{d}{dt}f(t) \not= 0$ for any $t$ in its domain. In particular, $\frac{d}{dt}\phi \circ \psi \not = 0$.
With regard to the composition, letting $\tilde{\gamma}(u)=\gamma(\phi(u))$, and $u=\psi(t)$, we get: $$\hat{\gamma} = \tilde{\gamma}(\psi(t)) = \gamma(\phi(\psi(t))) = \gamma(\phi \circ \psi (t))$$