Why is $SU(2)$ diffeomorphic to $S^3$? As the title,my question is why $SU(2)$ is diffeomorphic to $S^3$ ? How can I structure a diffeomorphic map between $SU(2)$ and $S^3$?
Please help me,and give me more details!
Thank you very much!
 A: By definition, we have that  $\displaystyle SU(2)=\left\{\begin{pmatrix}a &-\overline{b}\\  b &\overline{a}\end{pmatrix}: a, b\in\mathbb{C}, |a|^2+|b|^2=1\right\}.$
Since $\mathbb{R}^4\cong\mathbb{C}^2 $, we may think of $S^3$ as $\displaystyle S^3=\{(a,b)\in\mathbb{C}^2: |a|^2+|b|^2=1\}$.
Define a map from $S^3$to $SU(2)$ as:
\begin{aligned}F: S^3&\to SU(2)\\  F(a,b)&=\begin{pmatrix}a &-\overline{b}\\  b &\overline{a}\end{pmatrix}.  \end{aligned}
Can you show that $F$ is well defined, $F$ is injective, and $F$ is surjective (i.e. an isomorphism)? After that, you should show that $F$ and $F^{-1}$ are smooth (or $C^{\infty}$). (Hint: Show $\displaystyle SU(2)$ is a submanifold of of the set of 2 x 2 matrices with complex entries, or better yet $\mathbb{R}^8$. What can you say about $F$ and its inverse restricted to these submanifolds?)
A: Every $SU(2)$ matrix is a linear combination of Pauli matrices $\sigma_1, \sigma_2, \sigma_3 $ and the unit matrix I, which is a base of the associated Lie algebra of the SU2 group.
So every $SU(2)$ matrix  $X$ can be written a $X=x_1\sigma_1 +x_2\sigma_2 +x_3\sigma_3 +x_4.I$ and the condition that $det{X} = 1$ gives $x_1^2 + x_2^2 +x_3^2 +x_4^2 = 1$ which is the $S^3$ sphere. And every point on that sphere represents an $SU(2)$ matrix
