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).

  • $\begingroup$ 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"? $\endgroup$ – user30035 Mar 19 '13 at 21:26
  • $\begingroup$ Yes, you're right, it was really really easy! $\endgroup$ – John Mar 19 '13 at 21:33
  • $\begingroup$ I learnt this trick in my UG diff geom class; it probably has a completely standard name but if so, it escapes me. $\endgroup$ – 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)

  • $\begingroup$ Can you please state a precise theorem? I don't think the answer is so easy. Note that the function starts from an open subset of R^n but the codomain is R^m, with possible m>n. Of course if f(U) admits a differential manifold structure the answer is yes, but in general I think the situation is more complicated. $\endgroup$ – John Mar 19 '13 at 21:07
  • $\begingroup$ I think its called Inverse function theorem in english $\endgroup$ – trew Mar 19 '13 at 21:12
  • $\begingroup$ Which version of the theorem are you using? Sorry to be pedantic, but can you write a precise reference? I know that "it is everywhere", but it always has some assumption that is not verified in my question. For example it is usually stated for functions R^n --> R^{n-p}. $\endgroup$ – John Mar 19 '13 at 21:15
  • $\begingroup$ en.wikipedia.org/wiki/Inverse_function_theorem the fact that the dimension must be the same follows from you assumption that Df(x):R^n -> R^m is invertible(so you can use linear algebra to see that n=m) $\endgroup$ – trew Mar 19 '13 at 21:17
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    $\begingroup$ I am very sorry I meant that D(f) is injective. I must be a little tired. Again, sorry. I edit the question, mentioning that your answer is to another question and it is, of course, correct. $\endgroup$ – John Mar 19 '13 at 21:18

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