Im working on a problem but I can't figure it out now.
The problem is : If F:R^3 -> R^3 is a diffeomorphism and M is a surface in R^3, prove that the image F(M) is also a surface in R^3.(Hint : If x is a patch in M, then the composite function F(x) is regular, since F(x)* = F* x_* (tangent map))
I was trying to solve it. We can think of a patch x in M. then F(x) is regular as hint says. Also x^(-1)F(-1) is continuous since F is a diffeomorphism. But here, I think, we need that F(x) is one-to-one. x is one-to-one but F we don't know. How should I go further from here?
(is it right that diffeomorphism and one-to-one correspondence is not something related?)
To show $N:=F(M)$ is also a smooth surface we must present a collection of charts $(U_i, \Phi_i)$ such that $N \subset \bigcup_i \Phi_i(U_i) $, where each $\Phi_i$ is a homeomorphism, for all $p \in M$, there exists $j$ such that $p \in \Phi_j(U_j)$ and each $ \Phi_j$ is an immersion. Define $\psi_i = F \circ \phi_i$ where $\{(U_i, \phi_i)\}$ is a collection of maps which make $M$ a smooth surface.
$\bullet$ Show that $\{(U_i, \psi_i)\}$ make $N$ into a smooth surface. Hint: By chain rule $D\psi_i = DF \circ D \phi_i$ and $F$ a diffeomorphism implies $DF$ is an isomorphism. Now use a fact from linear algebra;
Given $U, V,W$ are finite-dimensional vector spaces, if $T: V \to W$ is a linear isomorphism and $S:U \to V$ has full-rank i.e $\textbf{dim}(S(U)) = \textbf{dim}(U)$ then $\textbf{dim}(T \circ S) = \textbf{dim}(U)$. The conclusion of this shows that each $\psi_i$ is an immersion.