The problem: "If $F:R^3\rightarrow R^3$ is a diffeomorphism and M is a surface in $R^3$, prove that the image F(M) is also a surface. (Hint: If x is a patch in M, then the composite function F(x) is regular, since $(x)^*=F^*x*$.)
My attempt at a solution (I saw a related post but it answers in terms of homomorphisms, which I know nothing about):
Each point p in F(M) has a unique point q in M such that p=F(q). Around each point q there is a neighborhood that is the image of a proper patch x. Since diffeomorphisms are 1-1 and invertible, the composite map Fx is a proper patch. Since the image of x is a neighborhood of q in M, its image, under F, is in F(M). So there is a neighborhood around p such that F(M) can be expressed as the image of the patch Fx. Thus F(M) is a surface.
Have I omitted anything?