I'm trying to prove the following result:
If $f$ is a map between two regular surfaces $S_1$ and $S_2$, $p \in S_1$, and if the differential $df_p : T_p S_1 \to T_{f(p)} S_2$ is an isomorphism (where $T_p S_1$ denotes the tangent plane to $S_1$ at $p$), then $f$ is (locally) a diffeomorphism.
The solution brings back the map $f$ to a map $g$ between Euclidean spaces, in order to apply the inverse function theorem:
It chooses $\phi_1 : U_1 \to S_1$ and $\phi_2 : U_2 \to S_2$ (where $U_i \subseteq R^2$), which are charts for $S_1$ and $S_2$, and defines $g := \phi_2^{-1} \circ f \circ \phi_1 : U_1 \to U_2$. Then, all we have left to do is to show that $g$ is a diffeomorphism: set $q = \phi_1^{-1}(p)$, then:
$dg_q = {d\phi_2^{-1}}_{f(p)} \circ df_p \circ {d\phi_1}_q$ (if I'm correct). Now, $df_p$ is an isomorphism, and, by definition, a chart $\phi$ is a smooth map such that $d\phi$ is injective. However, the solution automatically jumps to saying that $dg_q$ is invertible, and I do not know how to verify that.
Because that would require first to make sure that ${d\phi_2^{-1}}_{f(p)}$ is injective ; but even then, injectiveness of $dg_q$ would only ensure that it's a bijection on some set, not on the whole domain?
So how do I make sure that the inverse function theorem applies? Thanks for your time, let me know if something is unclear.