How to see that SL(2,C) is simply connected? I started reading about Lie groups and right now I'm trying understand why $SL(2,\mathbb{C})$ is simply connected. I have shown that $SU(2)$, being diffeomorphic to $S^3$, is simply connected. So my first thought is to construct a deformation retract of $SL(2,\mathbb{C})$ onto $SU(2)$. 
We know that by polar decomposition, every element $g \in SL(2,\mathbb{C})$ can be written as $u_g.x_g,$ where $u_g \in SU(2)$ and $x_g$ is a positive semi-definite self-adjoint element of $SL(2,\mathbb{C})$. And since the polar decomposition is unique, we can immediately define our retraction map as follow $$r : SL(2,\mathbb{C}) \longrightarrow SU(2)$$ $$g \mapsto u_g$$ Let $\iota: SU(2) \hookrightarrow SL(2,\mathbb{C})$ be the inclusion map. Now I have trouble constructing a homotopy between $\iota \circ r$ and the identity map on $SL(2,\mathbb{C})$ . Any hint?
Another approach I have is to show that $SL(2,\mathbb{C})$ is diffeomorphic to $S^3 \times M$, where $M$ is a $3-$ dimensional smooth manifold defined by $\left \{ (x_1,x_2,x_3,x_4)\in \mathbb{R}^4: x_1^2-x_2^2-x_3^2-x_4^2=1 \right\}$. Again constructing the diffeomorphism is easy, taking advantage of the uniqueness of polar decomposition of $g\in SL(2,\mathbb{C})$. So to show that $SL(2,\mathbb{C})$ is simply connected we now just need to show that $M$ is simply connected. But I don't see any easy way to prove this is true!
Anyways, to summarize, I have 2 questions:


*

*How to show $\iota \circ r$ is homotopic to the identity map on $SL(2,\mathbb{C})$?

*Why is $M$ simply connected?


Any insight/hint would be greatly appreciated!
 A: On 1) What you want to do is use Gram-Schmidt to do this. Consider matrices $(a_1, a_2)$, where the $a_i$ are column vectors. The Gram-Schmidt process replaces this matrix with $\left(u_1 := a_1/\|a_1\|, \frac{a_2-\text{proj}_{u_1}a_2}{\|a_2-\text{proj}_{u_1}a_2\|}\right)$. This is a retract $SL_2(\mathbb C) \to SU(2)$. We want to replace this with a deformation retract. So let's do each of these steps continuously; first consider $f_t(a_1, a_2) = (a_1/\|a_1\|^t, a_2\cdot \|a_1\|^t)$. This remains in $SL_2(\mathbb C)$ for all $0 \leq t \leq 1$, and $f_1(a_1,a_2)$ has a unit vector in the first column. Now consider $g_t(a_1, a_2) = (a_1, a_2-t\text{proj}_{a_1} a_2)$. Because we're just subtracting a scalar multiple of the first column this is again in $SL_2(\mathbb C)$ for all $t$. Concatenating these two maps gives you a homotopy from the identity to the Gram-Schmidt retract $SL_2(\mathbb C) \to SU(2)$.
This is true in absurd generality. First of all, the same argument applies to show that $GL_n(\mathbb C)$ deformation retracts onto $U(n)$, and $SL_n(\mathbb C)$ to $SU(n)$, and the same is true of their real counterparts. But we can do much better than that: the Malcev-Iwasawa theorem says that for any locally compact group $G$, $G$ deformation retracts onto its maximal compact subgroup $K$ (which exists and is unique up to conjugacy). Indeed there is some $n \in \mathbb N$ such that $G$ is homeomorphic to $K \times \mathbb R^n$. This includes, in particular, all Lie groups; I seem to remember that in the Lie group case you're actually diffeomorphic to $K \times \mathbb R^n$ but don't remember a reference.
On your second approach: the subset $M = \{x \in \mathbb R^4 \mid x_1^2 - x_2^2-x_3^3 - x_4^2 = 1,  x_1>0\}$ is diffeomorphic to $\mathbb R^3$, the map being $\mathbb R^3 \to M, x \mapsto (\sqrt{1+\|x\|^2},x)$. These are known as Weierstrass coordinates. Indeed your $M$ is better known as the hyperboloid model of hyperbolic 3-space; if you put the right semi-Riemannian metric on $\mathbb R^4$, then restricting it to $M$ gives you the hyperbolic metric. 
Either approach successfully shows that $SL_2(\mathbb C)$ is simply connected.
