Global invertibility of a map $\mathbb{R}^n\to \mathbb{R}^n$ from everywhere local invertibility 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.
 A: 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. 
A: 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
  
  
*
  
*$f$ is proper
  
*The Jacobian of $f$ is everywhere invertible
  
*$M_2$ 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.
