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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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It's not injective, as $h(1,x)=x_0$ for all $x\in S^{n-1}$. Perhaps you wanted $(0,1)$ or $[0,1)$ instead of $[0,1]$? –  Zev Chonoles Jan 19 '12 at 18:20
    
You're absolutely correct. I forgot to tell the whole story. I want to show that the map induced by the quotient map $q:[0,1]\times S^{n-1}\to [[0,1]\times S^{n-1}]/[\{1\}\times S^{n-1}]$ is surjective. So because of this I need only consider $t\in [0,1)$, as you suggested. –  Kyle Schlitt Jan 19 '12 at 18:24
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My usual suggestion in these situations is that you do write out the two pages of special cases. Once you have done it, you will be in a better position to try to reorganize the different cases and "compress" the argument. Elegance of argument is irrelevant when you do not have a complete, correct argument. So get one first and then worry. (In my experience, when a student tells me that he wants an elegant way to avoid the 5 pages of special cases he has in mind, the reality is that he does not have anything not even close to a complete argument!) –  Mariano Suárez-Alvarez Jan 19 '12 at 18:53
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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. –  user17786 Jan 19 '12 at 19:35

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