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For a vector $x_0$ interior to $B^n$, I have the function $h:[0,1)\times S^{n-1}\to B^{n}$ defined by $h(t,x) = tx_{0} + (1-t)x$.

I need to show it is injective but every time I start in the standard way (assuming $h(t_1, x_1) = h(t_2, x_2)$, it seems that I have to consider so many special cases. e.g. $x_{0}$ is parallel to $x_1$ but not to $x_2$, etc. And the whole argument turns into an absolute mess. In particular, the case where all three vectors are parallel is a mess. It seems that in itself it requires its OWN special sub-cases to be considered...

Is there an elegant way to justify this? I don't want to go on for 2 pages considering special cases because this is a very small part of a much bigger problem.

Any advice? Thank you!

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    Another way of proving it (that might enlight the way you have chosen): Try to prove that it defines a bijection onto its image (that is: calculate the image -that should be easy- and then construct a inverse map -it doesn't have to be continuous, although in this case it is). That would imply the map in injective.2012-01-19

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