Homeomorphic mapping The so-called Homeomorphic mapping theorem is that

If $H(z):\mathbb{R}^n\to\mathbb{R}^n$ is a continuous map and satisfies the
  following conditions:
(i) $H(z)$ is injective on $\mathbb{R}^n$,
(ii) $\lim_{\|z\|\to\infty}\|H(z)\|=\infty$,
then $H(z)$ is a homeomorphism of $\mathbb{R}^n$ onto itself.

How to prove the theorem? Please give the proof or recommend a book about the theorem. Thank you very much.
 A: I've never heard that name before, but I think the result is easy enough to prove.
The key result for this proof is

Invariance of domain: if $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ is continuous and injective, then $f$ is open and is a homeomorphism from $\mathbb{R}^n$ to $im(f)$.

(Actually IoD says more than this, but that's what I'll use here. Note that the statement "$f$ is a homeomorphism to its image" already follows from $f$ being an open map.)
So surjectivity is all that's left to prove. Suppose $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ is continuous, injective, and not surjective. Let $x$ be a boundary point of $\mathbb{R}^n\setminus im(f)$; since the latter set is closed ($f$ being open), $x\not\in im(f)$. 
Consider a sequence $y_i\rightarrow x$ with $y_i\in im(f)$ (such a sequence exists since $x$ is on the boundary of $\mathbb{R}^n\setminus im(f)$). Let $a_i=f^{-1}(y_i)$.
Now, think about the $a_i$s. Could they have bounded norm in $\mathbb{R}^n$? Well, then they'd all lie in some finite-radius closed ball, which is compact, so we would have a convergent subsequence $b_i\rightarrow b$. But the continuity of $f$ would then require $f(b)=x$, a contradiction.
So the $a_i$s don't have bounded norm. Then we can extract a subsequence $c_i$ whose norm tends to infinity. Now apply the condition on what $f$ does to norms . . .
A: The conditions imply that the map extends continuously to the one-point compactification.
So we have $\widetilde{f}: S^n \to S^n$. This extension is clearly injective by definition. Suppose $\widetilde{f}$ is not surjective. Therefore, it misses a point. It would follow that you would be able to embed $S^n$ in $\mathbb{R}^n$, which is absurd, by invariance of domain.
The extension being now bijective, continuous, and defined in compact Hausdorff spaces, you have a homeomorphism. Restrict to get back the original function.
