Proving injectivity of a function I have:
$$\begin{cases}  \rho\sin \theta \cos \phi= \rho'\sin \theta' \cos \phi' \\ \rho\sin \theta \sin\phi=\rho'\sin \theta' \sin \phi' \\ \rho\cos\theta=\rho'\cos \theta'\end{cases}$$
How do I show that $(\rho, \theta, \phi)= (\rho', \theta', \phi')$
 A: You certainly need some conditions on $\rho$, $\theta$ and $\phi$. It is standard to assume that $\rho\ge0$, $0\le\theta\le\pi$ and, say, $-\pi\le\phi\le\pi$, with some boundary conventions. I'll assume we are in a typical (non-boundary) point.
First, square the first two equations and add. We get on the left hand side $\rho^2\sin^2\theta(\sin^2\phi+\cos^2\phi)=\rho^2\sin^2\theta$, using that $\cos^2\alpha+\sin^2\alpha=1$ for all $\alpha$. The same way, the right hand side gives us $(\rho')^2\sin^2(\theta')$, so 
 $$ \rho^2\sin^2\theta=(\rho')^2\sin^2(\theta'). $$
Square the third equation and add, to get now on the left $\rho^2(\sin^2\theta+\cos^2\theta)=\rho^2$. Similarly, on the right we now get $(\rho')^2$, so $\rho^2=(\rho')^2$
and our convention that $\rho,\rho'$ are non-negative gives us $\rho=\rho'$. Assume now $\rho>0$.
From the sum of the squares of the first two equations, we now get $\sin^2\theta=\sin^2\theta'$, and our convention that $\theta,\theta'\in[0,\pi]$ gives us that $$\sin\theta=\sin\theta',$$ this is because $\sin\theta,\sin\theta'\ge0$ under this restriction. The third equation gives us $$\cos\theta=\cos\theta',$$ and from these two displayed conditions we now get $\theta=\theta'$.  
Now, from the first two equations we get $\sin\phi=\sin\phi'$ and $\cos\phi=\cos\phi'$, so also $\phi=\phi'$.  
