Invertibility of a function (Surjective/everywhere frechet)

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.

$$F(x)=\left( \begin{matrix} e^x \cos(y) \\ e^x \sin(y) \end{matrix} \right),$$ 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).
• If the question wanted to get the reader to appreciate the fact that periodic functions don't have a global inverse, then why not give us that $f$ is continuously differentiable too? – Displayname Feb 14 at 12:37