What is a parametrization and how can I parametrize an entire sphere? I have a question that asks to show that $S^2 = \{(x,y,z) \in \mathbb{R}^3|x^2+y^2+z^2=1\}$ is a differentiable manifold. My professor says that one way to do this is to define the following 6 parametrizations of the sphere, which cover the entire sphere.
$\vec{\phi_{i}}:V \to \mathbb{R}^3$ where $V = \{(u,v) \in \mathbb{R}^2|u^2+v^2<1\}$
$\vec{\phi_{1}}(u,v) = (u,v,\sqrt{1-u^2-v^2}) \qquad (z>0)$
$\vec{\phi_{2}}(u,v) = (u,v,-\sqrt{1-u^2-v^2}) \qquad (z<0)$
$\vec{\phi_{3}}(u,v) = (u,\sqrt{1-u^2-v^2},v) \qquad (y>0)$
$\vec{\phi_{4}}(u,v) = (u,-\sqrt{1-u^2-v^2},v) \qquad (y<0)$
$\vec{\phi_{5}}(u,v) = (\sqrt{1-u^2-v^2},u,v) \qquad (x>0)$
$\vec{\phi_{6}}(u,v) = (-\sqrt{1-u^2-v^2},u,v) \qquad (x<0)$
I don't understand what these parameterizations mean at all and I don't understand what a parameterization is. From what I can read online, it's some function but I'm not sure why this specific function with $u$ and $v$ is what we're using to cover the entire sphere. Can someone explain this to me please?
 A: For example the first gives $u^2+v^2+\left(\sqrt{1-u^2-v^2}\right)^2=1$.
Also there is $x=\sin\alpha$, $y=\cos\alpha\sin\beta$, $z=\cos\alpha\cos\beta$, where $\alpha\in[-\frac{\pi}{2},\frac{\pi}{2}]$ and $\beta\in[0,2\pi]$.
A: I don't understand what these parameterizations mean...
For all $(u,v)\in V:$
$$\vec{\phi_1}(u,v) = \left(u,v,\sqrt{1-u^2-v^2}\right)\in S^2$$
because
$$u^2 + v^2 + \left(\sqrt{1-u^2-v^2}\right)^2 = u^2 + v^2 + 1-u^2-v^2 = 1.$$
And the same for the other $\vec{\phi_i}$.
Also, every point of $S^2$ is equal to $\vec{\phi_i}(u,v)$ for some  $i$ and some $(u,v)\in V$.
I don't understand what a parameterization is...
What definition you know?
A: ${\bf Parametrization}$ :
Imagine that a point is moving on the plane, you can see the curve $\gamma \subset \mathbb{R}^2$ described by this point as a trajectory of a point in the plane when the time varies. Hence this curve can be seen as a vector function 
$X \; : \; [t_0 \; ; \; t_1] \longrightarrow \mathbb{R}^2$, 
which associates to any $t$, the vector $X(t) := (x(t),y(t))$, the position of the point at time $t$. We say that we parametrized the curve by the variable $t \in [t_0 \; ; \; t_1]$. The idea is that the curve is an object of dimension 1 even if it "lives" in a space of dimension great than 1 ! So we can describe it with ${\bf ONE}$ variable. 
For example, we can parametrize the circle of radius $R>0$ and centred at the origine by 
$X \; : \; [0 \; ; \; 2\pi[ \longrightarrow \mathbb{R}^2$, which associates to any $t$, the vector $X(t) := (\cos(t),\sin(t))$. 
You present the same thing with an object of dimension 2 living in 
$\mathbb{R}^3$, the sphere $S^2$, so you need ${\bf TWO}$ variables $(u,v) \in V$ to describe your sphere.
For $\Phi_1$ for example, if $(u,v)$ describes $V$ then the point $(u,v,+\sqrt{1-(u^2+v^2)} \in \mathbb{R}^3$ will describe the (open) upper half of the sphere $S^2$. Indeed, $u^2 + v^2 + \left(\sqrt{1-(u^2+v^2)}\right)^2 = 1$, and you know I suppose the cartesian equation of the sphere $S^2$. The same thing is done to try to cover all the sphere $S^2$ by open 
halfs of the sphere : you need 6 $\Phi_i$'s to cover all the sphere.
On the other hand, to prove that the parametrization you gave implies that the sphere is a differential manifold, you have to use a lemma of differential geometry.
