$SU(n)$ is simply connected (proof without fibrations, $n>2$)

Question

How to show that $SU(n)$ is simply connected for $n>2$ if I don't know about fibrations yet? For $SU(2) \cong S^3$ the fact is said to be known. For any matrix $A \in SU(n)$ there is a matrix $S \in U(n)$ such that $$ A = S^{-1}DS, $$ where D has a form $$ D = \begin{bmatrix} e^{i \varphi_1} &0&0 & \cdots &0 \\ 0 & e^{i\varphi_2} & 0 & \cdots & 0 \\ 0 & 0 & e^{i\varphi_3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \\ 0 & 0 & 0 & \cdots & e^{i\varphi_n} \end{bmatrix} $$ and since $\det D =1$ we have $\varphi_1+\ldots+\varphi_n = 0 \mod 2\pi$. Maybe one can find some deformation retract of $SU(n)$ with easy computable fundamental group (which will be trivial)?

Answer 1

Perhaps you might try thinking about it starting with $SU(2)$. What is this homeomorphic to?

Answer 2

Using Morse theory (I will let you fill in the details):

Consider the Morse function $f:SU(n)\to\mathbb{R}$ given by $[z_{ij}]\mapsto\text{Re}(\sum_ic_iz_{ii})$ for fixed constants $1<c_1<<c_2<\ldots<c_n\in\mathbb{R}$. It has a critical point of index 0, and the next smallest index is 3 (figure out why we want $c_1<<c_2$). Thus $SU(n)$ has no 1-handles, meaning it is simply-connected.

For $n=2$, it has only two critical points $\lbrace\pm\text{id}\rbrace$. In particular, $SU(2)$ is built up by two 3-disks and hence is diffeomorphic to $S^3$.