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I would like to know what are the methods people have used to prove that a topologically (or analytically) nice mapping $f: B\to \Omega$ is injective? Above, $B$ is the unit ball in $\Bbb R^n$ and $\Omega$ is a "nice" domain in $\Bbb R^n$, $n>2$. I used the words topologically nice or analytically nice since I do not know the exact condition that will imply the injectivity.

For topologically nice map, we may assume it is continuous, discrete and open, proper, ... and so on. For analytically nice, we may assume nice regularity. For the question above, one can also add assumptions on the target domain $\Omega$. For instance, if $\Omega$ is also a (unit) ball, and if the mapping $f$ has compact branch set (which is the set of points in the source such that $f$ is not locally injective at that point), then $f$ is injective. This can be seen in the following way, the closeness (or properness) of the mapping implies that the boundary get mapped to the boundary and so we may choose a simply connected neighborhood $U$ around the boundary of the image and analyze the preimages of $U$. Since it does not touch $f(B_f)$, it is easy to show that $f$ forms a covering map between these neighborhoods. Since the target $U$ is simply connected, we conclude that $f$ is one-to-one when restricted to $f^{U}$. On the other hand, the degree of $f$ is constant in the ball and so we conclude that $f$ is one-to-one.

It would be greatly appreciated if one can tell me some other methods or conditions that sufficient to conclude that a mapping is injective.

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To Henry T. Horton: thank you very much for editing. Why people cannot use latex command directly in text. I always feel difficult to write down a mathematical formula here. – Changyu Guo Nov 27 '13 at 9:21
I do not understand your claim about branch set: Contract a compact interval in the open 3-ball to a point. The result is still diffeomorphic to the 3-ball; using this construction you can get a smooth map $B^3\to B^3$ which is not locally injective on a nonempty compact subset. Do you assume that the map is light? I do not believe this is enough either (in your argument, there is no way you can prove that the map is a covering near the boundary, it is only locally injective). – Misha Nov 27 '13 at 11:16
To Misha: I have assumed that f is continuous, discrete and open, closed and the branch set is compact in the unit ball B. So in particular, f is a covering mapping when restricted to B-f^{-1}(f(B_f)). – Changyu Guo Nov 27 '13 at 12:46

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