Let $U,V \subseteq \mathbb{R^n}$ be open. Suppose that $f: U \rightarrow V$ is surjective and is everywhere Frechet differentiable with $df(x)$ invertible for all $x \in U$. Is $f$ invertible?
I know that one of the conditions for the inverse function theorem is that $f$ is continuously differentiable, which we don't necessarily have. If the question didn't give us that $f$ was surjective, I'd have concluded that $f$ doesn't have to be invertible, but I'm just wondering what the question is trying to tell us by mentioning surjectivity of $f$, if it implies that $f$ is continuously differentiable I don't see how.
No, and the issue is not the continuity of the derivative. To see this, you could take the function from wikipedia's article on the inverse function theorem:
$$F(x)=\left( \begin{matrix} e^x \cos(y) \\ e^x \sin(y) \end{matrix} \right),$$ which is a surjective continuously differentiable function from $\mathbf{R}^2$ to $\mathbf{R}^2 \setminus \{0 \}$ whose differential is everywhere invertible. This has an inverse everywhere locally, but no global inverse since it is periodic in $y$.
(This is really the holomorphic function $z \mapsto e^z$ from $\mathbf{C}$ to $\mathbf{C}^\times$ in disguise).