Show $T(u,v,w) = (u \cos v \cos w, u \sin v \cos w, u \sin w)$ is onto the unit sphere 
$T: \mathbb{R^3} \rightarrow \mathbb{R^3}$ is defined by $T(u,v,w) = (u \cos v \cos w, u \sin v \cos w, u \sin w)$. Show that $T$ is onto the unit sphere, $x^2 + y^2 + z^2 = 1$.

I believe I have to show:
$$
u \cos v \cos w = x \\
u \sin v \cos w = y \\
u \sin w = z
$$
For $x^2 + y^2 + z^2 = 1$. Does that mean I have to solve for $u, v,$ and $w$? How can I do this? And how do I make my codomain restricted to the unit circle instead of $\mathbb{R^3}$?
 A: Well first, the function is defined with a codomain of $\mathbb{R}^3$ so the question isn't asking to show that $T$ itself is onto, but that $\forall \; \vec{v} \in B_1(0) \; \exists \vec{w}$ s.t. $T(\vec{w}) = \vec{v}$ where $B_1(0)$ is the unit sphere.
In order to do this we need to take some arbitrary element of $B_1(0)$, meaning:
$$
\text{Let } x,y,z \in \mathbb{R} \text{ s.t. } x^2 + y^2 + z^2 = 1
$$
and show that $\exists \; u,v,w$ s.t. $T(u,v,w) = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$
If we solve the system of equations you have above:
$$
u \cos v \cos w = x \\
u \sin v \cos w = y \\
u \sin w = z
$$
with the property that $x^2 + y^2 + z^2 = 1$. First we can see that in order to have the property that $x^2 + y^2 + z^2 = 1$ we must have that
$$
1 = u^2 \cos^2 v \cos^2 w + u^2 \sin^2 v \cos^2 w + u^2 \sin^2 w = u^2 \left(\cos^2 w \left( \cos^2 v + \sin^2 v \right) + \sin^2 w \right) = u^2
$$
so that $u = \pm 1$. Continuing, for convenience let's choose $u = 1$ (note in general you cannot simply choose this, why?). So now we need to solve the system:
$$
\cos v \cos w = x \\
\sin v \cos w = y \\
\sin w = z
$$
Now we get from the last equation that $w = \arcsin z$ and find the following equations:
$$
\cos v \cos \arcsin z = x \\
\sin v \cos \arcsin z = y
$$
We can divide these equations to see that
$$
\tan v = \frac{y}{x}
$$
so that $v = \arctan \frac{y}{x}$. Now put
$$
\vec{w} = \begin{bmatrix} 1 \\ \arctan y / x \\ \arcsin z \end{bmatrix}
$$
so that
$$
T(\vec{w}) = \begin{bmatrix} x \\ y \\ z \end{bmatrix}
$$
And thus $T$ is onto the unit sphere, since $x,y,z$ were arbitrary. Hopefully this helps!
