Let $S$ be a regular surface and let $F$ be a diffeomorphism. How can I prove that the image $F(S)$ is a regular surface ?
Consider a regular surface $S \subset \mathbb R^3$ and a diffeomorphism $F: \mathbb R^3 \rightarrow \mathbb R^3$.
How can I prove that the image $F(S) \subset \mathbb R^3$ is a regular surface ?
My idea is:
Given $x \in F(S)$, I look at $y = F^{-1}(x) \in S$. By definition there exist open sets $U \in \mathbb R^2$ and $W \in \mathbb R^3$ such that $x \in S \cap W$ and $\sigma: U \rightarrow S \cap W$ a homeomorphism that is smooth (and some more).
I then look at $F(S \cap W) = F(S) \cap F(W)$, but how do I know $F(W)$ is open ? Should I maybe look at the composition of functions ?
How can this be used to prove that the ellipsoid $$\{ (x,y,z) \in \mathbb R^3 \mid \frac {x^2} {a^2} + \frac {y^2} {b^2} + \frac {z^2} {c^2} = 1\}$$ is a regular surface ??
You have the right idea to look at the preimage and compose the charts with the diffeomorphism. To proceed, remember that diffeomorphism is, in particular, a homeomorphism, and so identifies open sets in the domain and the range.
In considering the ellipsoid, is there something similar to an ellipsoid you already know is a regular surface? Can you relate this something to the ellipsoid via a map which is a diffeomorphism of $\mathbb{R}^3$?