The diffeomorphism of $\mathtt R^n$ [duplicate]

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Let $f:{\mathtt R^n} \to {\mathtt R^n}$ be a smooth map and $df(x)$ is nonsingular for all $x$, in addition, $\mathop {\lim }\limits_{\left| x \right| \to + \infty } \left| {f(x)} \right| = + \infty$, then is it necessarily that $f$ is injective?

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marked as duplicate by t.b., Zhen Lin, Davide Giraudo, joriki, SrivatsanOct 29 '11 at 16:58

As explained in the thread above, even more is true: from your hypotheses it follows that $f$ is a diffeomorphism (this is called the Hadamard-Cacciopoli (inverse function) theorem). –  t.b. Oct 29 '11 at 16:12
Because $f$ is proper and locally diffeomorphic, $f: \mathtt {R^n} →\mathtt {R^n}$ is an universal covering map. Since $\mathtt {R^n}$ is simply-connected, then the deck transformation group is trivial and therefore $f$ is injective.