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Is transformation $g:x=(x_1,x_2,x_3)\mapsto \frac{x}{\|x\|}$ injective? What if $x_1=1$?

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It's not injective because for any $x\in\mathbb{R}^3$, $\lambda >0$, you have $g(x)=g(\lambda x)$. Conversely,

$$g(x)=g(y)\Rightarrow \exists \lambda >0,\, y=\lambda x$$

Now if you fix $x_1 =1$, it becomes injective because it gives you $1=\lambda *1$. Then $\lambda=1$ and $x=y$.

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