what does "SU2 is homeomorphic to the unit 3-sphere in R4 " mean Michael Artin's book on algebra
here is a section starting from " $SU_2$  is homeomorphic to the unit 3-sphere  in $R^4$ " and then   compare the $SU_2$ to the unit 3-sphere $x_1^2 + x_2^2 + x_3^2 + x_4^2 = 1 $ I am a bit confused


*

*this unit-3 sphere can't be drawn right? this is a super gemoetry object in my understanding

*what does the comparison longtitude and latitude to conjugacy classes of $SU_2$ mean on earth

*What does homeomorphic mean ? I cannot think of it as isomorphic or homomorphic right ?


my understanding here is that by thinking concretely in 3-dimenisonal space, we can understand the structure of $SU_2$ better. can someone help elaborate a little bit ?
 A: Let me give you one reason why we want to think of $\textrm{SU}(2)$ as the unit sphere: A priori, $\textrm{SU}(2)$ is only a closed subgroup of $\textrm{GL}_2(\Bbb{C})$. Now it is a topological group (specifically a matrix Lie group), meaning to say in the simplest terms it is a topological space and a group at the same time. Now suppose I want to understand topological properties of $\textrm{SU}(2)$, namely whether or not it is connected/path connected/compact, etc. 
Now all these things are not easy to see if you look at $\textrm{SU}(2)$ as just a bunch of matrices however: upon identifying it as a topological space with $S^3$, all these properties become obvious immediately. The 3-sphere is connected, path connected and is compact.
Something else that we get from this identification is we know that $S^n$ for $n > 1$ is simply connected. In simplest terms, any loop in your space at a point $x_0$ can be deformed continuously to the point $x_0$. It follows that $\textrm{SU}(2)$ is simply connected by it being homeomorphic to $S^3$. In algebraic topology, we say that
$$\pi_1(\textrm{SU}(2),x_0) = 0$$
and this is something very powerful: We already have a covering map $\Phi : \textrm{SU}(2) \longrightarrow \textrm{SO}(3)$ that I have defined here. The fact that the fundamental group of $\textrm{SU}(2)$ is trivial tells us immediately that it is the universal cover of $\textrm{SO}(3)$. Once we know the existence of this universal cover, in turn we can classify all up to isomorphism all non-based covering spaces of $\textrm{SO}(3)$.
And all of this came from just looking at $\textrm{SU}(2)$ as the 3-sphere!
A: The $ 2 \times 2$ complex unitary matrices of determinant $1$ are the elements of the set 
$$\left\{ \left( \begin{array}{clcr} z & w \\ -\overline{w} & \overline{z} \end{array} \right) :
z, w \in \mathbb{C}, |z|^{2} + |w|^{2} = 1 \right\}.$$ Let $M(z,w)$ denote the matrix specified by the choice of two such complex numbers $z,w$ with $|z|^{2}+ |w|^{2} = 1, $ say $z = x+iy$ and $w = u + iv$ . One choice of homeomorphism is 
$M(x+iy,u+iv) \to (x,y,u,v).$
A: As for (1) you can visualize it as follows: First take your 2-sphere... that 2-sphere is nothing more than a union of circles (1-spheres) stacked together with varying radii (radii goes from 0 to 1 to 0).  This holds similarly for the 1-sphere as a collection of 0-spheres (pairs of points of varying separation). Thus view the 3-sphere as a collection of 2-spheres of varying radii -- at the ends you have a point, and at the center you have a standard unit 2-sphere.
