I have some problems to prove that $SU(n)$ is simply-connected, for $U(n) = \{ M \in GL_n(C) \; | \; ^{t}\overline{M}M = Id \;\}$. In fact, there is some indications (to follow if it's possible). First, let's define $N=(0, \dots, 1)$ the North pole of $S^{2n-1} \subset \mathbb{C}$. We suppose $n \geq 2$. Then :
Let $e_N : SU(n) \rightarrow \mathbb{S}^{2n-1}$ the application which evaluates the matrix at the $N$, i.e : $e_N(A) = A \cdot N$. Prove that it exists a continuous section : $s : \mathbb{S}^{2n-1} -{N} \rightarrow SU(n)$ of $s$, i.e $e_N \circ s = Id$. Then, show by induction that $SU(n)$ is simply connected, using Van Kampen.
For the first point, so to find the section, let's $z \in \mathbb{S}^{2n-1} -N$. Then, we want to say that $s(z)(N) = z$ and $s(z) = Id$ over $Vect(N,z)^{\perp}$. Then, we have defined $s$ over $N$ and $Vect(N, z)$, and as we should send an orthonormal basis on an orthonormal basis, and as $dim(Vect(N, z)^{\perp} + Vect(N))=2n-1$, finally $s$ is totally determined.
Then, I don't succeed to do the second part, i.e showing that $SU(n)$ is simply connected. Actually, let's suppose the result is true for $SU(n-1)$. We can identify $SU(n-1)$ with $SU(Vect(N)^{\perp}) \subset SU(n)$ (by the "natural" inclusion). Then, I would like to consider : $S^{2n-1} - N \times SU(n-1)$ and $S^{2n-1} - S \times SU(n-1)$ (actually, the matrix associated, the first column would be the vector of $S^{2n-1} - N$ and the others columns would be the one of the element of $SU(n-1)$) because each factor are simply connected, but it's not a open cover of $SU(n)$, as it's not even elements of $SU(n)$. So, it should be here that we have to used the section.
But I don't figure out how to do it. Someone could help me ?
Thank you very much !
The group $SU_{n+1}$ acts transitively on $S^{2n+1}$ and the subgroup stabilising the basepoint $N=e_1$ is $SU_n$. Since $SU_n\subseteq SU_{n+1}$ is a closed Lie subgroup the projection to the quotient
$$p:SU_{n+1}\rightarrow SU_{n+1}\rightarrow SU_{n+1}/SU_n\cong S^{2n+1}$$
admits local sections and so has the structures a locally trivial $SU_n$-bundle.
Since $U_N:=S^{2n+1}\setminus N$ is contractible any local section defined over an open subset of it extends to a section $s$ define on all of $U_N$, and this scetion gives rise to an isomorphism
$$SU_{n+1}|_{U_N}\cong U_N\times SU_n$$
of principal $SU_n$-bundles over $U_N$ in the standard way. There is of course a similar isomorphism
$$SU_{n+1}|_{U_S}\cong U_S\times SU_n$$
over the open subset $U_S:=S^{2n+1}\setminus S$.
Over the intersection $U_{NS}:=(S^{2n+1}\setminus N)\cap(S^{2n+1}\setminus S)\cong S^{2n}\times(-1,1)$ we have therefore $$SU_{n+1}|_{U_{NS}}\cong U_{NS}\times SU_n.$$
Note then that $p^{-1}(U_N)=SU_{n+1}|_{U_N}$ and $p^{-1}(U_S)=SU_{n+1}|_{U_S}$ are open and cover $SU_{n+1}$ Hence we have a pushout square of open subsets $\require{AMScd}$ \begin{CD} p^{-1}(U_N)\cap p^{-1}(U_S)=p^{-1}(U_N\cap U_S)@>>> p^{-1}(U_N)\\ @VV V @VV V\\ p^{-1}(U_S) @>>> SU_{n+1}. \end{CD}
Given the bundle isomorphisms above this square becomes $\require{AMScd}$ \begin{CD} U_{NS}\times SU_n@>>>U_N\times SU_n\\ @VV V @VV V\\ U_S\times SU_n @>>> SU_{n+1} \end{CD} and finally if we replace everything up to homotopy (which is all we need to apply Van Kampen) we have that $SU_{n+1}$ is the homotopy pushout \begin{CD} S^{2n}\times SU_n@>>> SU_n\\ @VV V @VV V\\ SU_n @>>> SU_{n+1}. \end{CD} The process of producing pushout squares of this form is a general procedure called clutching, which applies to principal and vector bundles over spheres. You can check Hatcher's notes Vector Bundles & K-Theory for a detailed discussion.
If you're willing to trust me I will state without proof that you can arrange for the vertical left-hand map in the last diagram to be the projection onto the second factor, and the top horizontal map to have the form $(z,A)\mapsto \varphi(z)A$, where $\varphi:S^{2n}\rightarrow SU_n$ is a generator of $\pi_{2n}U_n\cong \mathbb{Z}_{n!}$ ($\varphi$ is the image of the identity under the connecting homomorphism $\Delta:\pi_{2n+1}S^{2n+1}\rightarrow \pi_{2n}U_n$ in the long exact sequence I mention in the comments).
Anyway, we don't need this, as $SU_1=\{1\}\subseteq U_1\cong S^1$ is just a point (also $SU_2\cong S^3$) so we can inductively apply van Kampen to any one of the previous diagrams to get the result that
$$\pi_1SU_n=0\qquad \forall n\geq 1.$$