Show that $\mathcal{M}(\mathrm{SU}(2))\cong S^{3}.$ $\mathcal{M}$ here is the underlying manifold of the group.
What I tried:
To show we need to establish a diffeomorphism between them.
The special unitary group $\mathrm{SU}(2)$ can be represented as a group of $2\times2$ complex matrices with determinant $1$ and unitary elements. Let $U$ be an element of $\mathrm{SU}(2)$, and it can be written as:
$ U = \begin{bmatrix} a & -\overline{b} \\ b & \overline{a} \end{bmatrix} $
where $a$ and $b$ are complex numbers satisfying $|a|^2 + |b|^2 = 1$ (due to the unitary condition).
Now, let's define a map $\phi: \mathrm{SU}(2) \rightarrow S^3$; $ \phi(U) = (a, b) \in \mathbb{C}^2 $.
This is where I'm stuck. Is this correct so far? How can I continue further?