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.

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    $\begingroup$ 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 $\mathbb{R}^n$. $\endgroup$ – t.b. May 26 '11 at 20:02

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.

  • $\begingroup$ 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. $\endgroup$ – Listing May 26 '11 at 21:03
  • $\begingroup$ 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 $\lim_{|x|\to\infty}|f(x)|=\infty$? $\endgroup$ – flonk Sep 19 '14 at 18:18
  • $\begingroup$ @flonk If your condition does not hold, there exists a sequence $x_n$ and $L>0$ with $|x_n|\to\infty$ and $|f(x_n)|\leq L$, thus the inverse of $B_L(0)$ under $f$ is not bounded and a fortiori not compact. Hence, $f$ is not proper. For the other direction, assume $f$ is not proper. Then there exists a compact set $C$ such that $f^{-1}(C)$ is not compact. If $f^{-1}(C)$ was bounded, each sequence would have a converging subsequence, and by continuity the limit of this sequence would be in $f^{-1}(C)$ too, contradicting non-compactness. Thus, the preimage of the bounded set $C$ is unbounded $\endgroup$ – Bananach Aug 13 '17 at 17:02
  • $\begingroup$ (1) Do you mean $l2$ norm with $|x|$ and $|f(x)|$? (2) Does the condition $\lim_{|x|\rightarrow \infty}|f(x)|=\infty$ need to be true for all $x$? $\endgroup$ – Alexander Mathiasen Sep 3 '20 at 9:25

Because $f$ is proper and locally diffeomorphic, $f:\mathbb R^n\to \mathbb R^n$ is an universal covering map. Since $\mathbb R^n$ is simply-connected, the deck transformation group is trivial and therefore $f$ is injective.

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


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