I'm trying to show that a generalized helicoid is a regular surface, namely that the trace of the function$$x(s,u) = (f(s) \cos u, f(s) \sin u, g(s) + cu),$$where $\delta(s) = (f(s), g(s))$, $s \in (a,b)$, is a regular curve, $f(s) > 0$ and $u \in (0, 2 \pi)$, is regular. We have to show that $x$ is differentiable, a homeomorphism and $dx_q$ is one-to-one for every $q$. First and last condition are easy to prove. A homeomorphism not so much. However, I came across the following result:
Let $x: U \subset \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be a regular parametrized surface and let $q \in U$. Then there exists a neighborhood $V$ of $q$ in $\mathbb{R}^2$ such that $x(V) \subset \mathbb{R}^3$ is a regular surface.
In the result above, a parametrized surface is a surface defined as the trace of map with one-to-one differential for every $q$, which is what the generalized helicoid is. So I guess my question is this: Is it not necessary to prove that $x$ is a homeomorphism in this particular case? I think the answer is yes, since (if I understood the result above correctly) we can guarantee the existence of a local frame around every point, but I'm still not completely sure.