Suppose we have a parametric curve in $\mathbb{R^2}$, i.e. $\phi: [a, b] \subset \mathbb{R} \to \mathbb{R}^2$, $\phi(s) = (\phi_1(s), \phi_2(s))$ for $s \in [a, b]$. Suppose there is a different parameterization of the curve, $\psi: [c, d] \in \mathbb{R} \to \mathbb{R}^2$, $\psi(t) = (\psi_1(t), \psi_2(t))$ for $t \in [c, d]$. If $\psi$ and $\phi$ are locally invertible, we define a reparameterization as $f: [a, b] \to [c, d]$, $f(s) = (\psi^{-1} \circ \phi)(s)$.
My question is whether, if we know that $\phi$ and $\psi$ are smooth, we can conclude that $f$ is also smooth. My professor said that we can't use the chain rule, because although we map from an open set in $[a, b]$, the part of the curve which we then map back to a part of $[c, d]$ is not itself an open set in $\mathbb{R}^2$. He then provided a simple argument by which we can extend the functions in question to map from open sets in $\mathbb{R}^2$ to other open sets in $\mathbb{R}^2$, allowing for use of the multivariable chain rule.
I would appreciate if someone could provide a precise statement of the multivariable chain rule which states that the images should be open sets. I can't find this anywhere.
Could someone provide an intuitive explanation, perhaps with a counterexample in an analogous situation where this doesn't hold, of why we can't conclude that the inverse of the smooth function $\psi$ is itself smooth?