# Showing a map is injective.

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.

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. –  roo Jan 19 '12 at 18:24