A condition of smooth homeomorphism of $\mathbb R^2$

30 Views Asked by At

Suppose that $f:\mathbb R^2\to\mathbb R^2$ is a $C^1$ function. If the Jacobian determinant $J(x,y)$ is not zero, $\forall x,y$ and $|f(x,y)|\to\infty$ as $|(x,y)|\to\infty$, then show that $f$ is a homeomorphism.(In fact, this is a necessary and sufficient condition and the converse is easier to be proved.)

I can prove it using some knowledge in topology, that is, $f$ is a proper local diffeomorphism and hence a covering(See http://www.math.ubc.ca/~oantolin/notes/propetale.html). By the connectness we can know that $f$ is an injection. The other details are easy to verify.

But I hope that anyone could give a more elementary proof, just using knowledge in mathematical analysis.