This is problem is from boothby's text on differentiable geometry:

Let $m \leq n$ and suppose $U$ be an open set in $\Bbb R^n$. Let $f: U \to \Bbb R^m$ a $C^{1}$ map which is also injective. Further suppose that $f^{-1}:f(U) \to U$ is also $C^{1}$. Then show that $m$ cannot be strictly less than $n$.

Here's what I've managed to deduce: the problem is trivial if one invokes brouwer's invariance of domain. But I don't want to use that. Also I do not wish to use the inverse function theorem.


1 Answer 1


I suppose this might work.Since $f^{-1}$ is differentiable from $f(U)$ to $U$, it is the restriction of a differentiable map $g$ defined on an open subset $V$ of $\mathbb{R}^{m}$ to $f(U)$.Of course, $f(U)$ is contained in $V$.Now consider $f$ from $U$ to $V$.Since $f(U)$ is in $V$, we have $g\circ f = id_{U}$. From chain rule, it follows that $df$ is injective, implying that $m=n$.

  • $\begingroup$ This is fine but where is the fact that f and f inverse are C1 being used? $\endgroup$ Commented May 15, 2020 at 15:54
  • $\begingroup$ Since chain rule holds for differentiable functions, we do not need to worry about continuity of derivatives at all here. Infact, this shows that the weaker assumption of differentiability suffices quite well,once you are willing to extend your concept of differentiability to domains which are not necessarily open,using the concept of restriction.This is the way it's defined in many standard texts,Spivak,for instance. $\endgroup$ Commented May 15, 2020 at 16:07
  • $\begingroup$ In many instances you will see the problem has been mentioned with conditions slightly stronger than those required.This is usually done with the intention to make the student apply some theorem to solve it, which might not necessarily hold under weaker assumptions. $\endgroup$ Commented May 15, 2020 at 16:10
  • $\begingroup$ Thanks this cleared up my doubts. $\endgroup$ Commented May 15, 2020 at 16:20

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