An injective smooth function with injective differential must have a continuous inverse?

Let $U \subset \mathbb R^n$ be an open subset and let $f \colon U \to \mathbb R^m$ be a $C^\infty$ function. We suppose that $f$ is injective and that the differential $Df(x)$ is injective for all $x \in U$. Does it follow that the inverse function $f^{-1} \colon f(U) \to U$ is continuous?

The question is motivated by the fact that some authors require continuity of the inverse in the definition of a parametrized surface in $\mathbb R^3$ and some authors does not.

I think the answer is "no", but I cannot find an example.

Note: the answer below by trew is to a previous question of the question, where I wrote "$D(f)$ invertible" by mistake (in which case trew's is of course correct).

• Now you've edited the question, how about $n=1$, $m=2$ and the map is from the open interval onto something which looks a bit like a letter "e"? – user30035 Mar 19 '13 at 21:26
• Yes, you're right, it was really really easy! – John Mar 19 '13 at 21:33
• I learnt this trick in my UG diff geom class; it probably has a completely standard name but if so, it escapes me. – user30035 Mar 19 '13 at 21:41

Yes this is true since your map is open by the local diffeomorphism theorem. We can find for every $x \in U$ an open $U_x$ such that f is a diffeomorphism from $U_x$ to $f(U_x)$.(i think we need that f is in $C^{1}$ not just differentiable)