Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was told by a tutor that if $f: \mathbb{R}^n \longrightarrow \mathbb{R}^n$ has an invertible Jacobian Matrix for all $x \in \mathbb{R}^n$ and $\lim_{|x_k| \rightarrow \infty}|f(x_k)|=\infty$ for all such sequences, then $f$ is already globally bijective.

This was very surprising for me (as this seems to be a very strong statement), so I tried to search on the internet for this theorem but I only came up with theorems about local inverses. I hope someone here can give me a reference or a name of this theorem so I could read a detailed proof. I sanity checked it for $n=1$ where it works perfectly.

share|cite|improve this question
Okay, surjectivity seems rather easy: the image is open by invertibility of the Jacobian and closed since for $y_{n} = f(x_{n}) \to y$, the sequence $x_{n}$ lies in a bounded set, hence it has a convergent sub-sequence, $x_{n_k} \to x$, and as $f$ is continuous we have $y = f(x)$, hence the image must be all of $\mathbb{R}^n$. – t.b. May 26 '11 at 20:02

2 Answers 2

up vote 20 down vote accepted

The result you ask about is called Hadamard's global inverse function theorem or sometimes Hadamard-Cacciopoli theorem. Googling these keywords reveals an entire industry of such local invertibility + something implies global bijectivity results.

Unfortunately, I was unable to find an accessible proof of this result. Among several sources I looked at, by far the best bet seems to be the presentation in Section 6.2 of the beautiful book by S.G. Krantz and H.R. Parks, The implicit function theorem: history, theory, and applications, Birkhäuser, 2002. The proof given there is essentially self-contained and doesn't assume much knowledge on the reader's side. Nevertheless, I should point out that the title of Chapter 6 is Advanced implicit function theorems, so it's definitely not for the faint-hearted.

In fact, a more general result is the following, also due to Jacques Hadamard. It is a bit, but not very much, harder to prove than the result you ask about.

If you don't know what a manifold is, simply replace $M_1$ and $M_2$ by $\mathbb{R}^n$ in the theorem below, and you obtain the result you're asking about — for $\mathbb{R}^n$ condition 3. is satisfied and condition 1. translates precisely to the condition $\lim\limits_{|x| \to \infty} |f(x)| = \infty$ your tutor told you.

Theorem (Hadamard)

Let $M_1, M_2$ be smooth and connected $n$-dimensional manifolds. Suppose $f: M_1 \to M_2$ is a $C^1$-function such that

  1. $f$ is proper
  2. The Jacobian of $f$ is everywhere invertible
  3. $M_2$ is simply connected.

Then $f$ is a homeomorphism (hence globally bijective).

So, as I said, this theorem is not trivial at all and both this and the result you're interested in can be found in the book I mentioned above. Quite a bit of googling didn't yield a simple(r) proof of the theorem you ask about, but as you have the key-words now, maybe you find something that suits you.

Added: I should have mentioned the better known Cartan-Hadamard theorem which is closely related but seems a bit more geometric in its nature.

share|cite|improve this answer
Thank you that was helpful. I thought it is easier to proof, but only the one direction you mentioned seems easy. Knowing the theorem I will surely find more references. – Listing May 26 '11 at 21:03
Thank you. For somebody with almost no background in topology, could you give a hint why for $M_2=\mathbb{R}^n$ properness translates to $\lim_{|x|\to\infty}|f(x)|=\infty$? – flonk Sep 19 '14 at 18:18

Because f is proper and locally diffeomorphic, f:Rn→Rn is an universal covering map. Since Rn is simply-connected, then the deck transformation group is trivial and therefore f is injective.

The same method can be applied to the general theorem of Hadamard

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.