Different ways of seeing how $\text{SL}(2, \mathbb{R})$ is not simply connected? As the question title suggests, could anybody sketch some of the different ways of seeing how the group $\text{SL}(2, \mathbb{R})$ is not simply connected?
 A: The standard way is to prove that $SO(2) \hookrightarrow SL_2(\Bbb R)$ is a homotopy equivalence. Think Gram-Schmidt. This has been detailed elsewhere plenty of times, so I won't go into it here.
Here's a mildly geometric proof. $SL_2(\Bbb R)$ acts on $\Bbb H$, considered via the unit disc model, by orientation preserving isometries; the only elements that act trivially are $\pm I$, leading to an action of $PSL_2(\Bbb R) = SL_2(\Bbb R)/\pm I$.. In particular we get an action of $PSL_2(\Bbb R)$ on its ideal boundary, $S^1$. There is a natural map $f: PSL_2(\Bbb R) \to S^1$, given by $\phi \mapsto \phi(x)$, where $x$ is a chosen basepoint. There is a copy of $S^1$ inside $PSL_2(\Bbb R)$ such that this map is an isomorphism (rotations of the hyperbolic plane). The loop this represents in $PSL_2(\Bbb R)$ cannot possibly be null-homotopic, for if it was, then composing the null-homotopy with $f$ would give a null-homotopy of the identity map $S^1 \to S^1$; indeed the same argument shows it must have infinite order in $\pi_1(PSL_2(\Bbb R))$. Its square lifts to $SL_2(\Bbb R)$, where it also must have infinite order. Thus we conclude.
Alternatively, note that $PSL_2(\Bbb R)$ acts transitively and freely on the unit tangent bundle $T^1\Bbb H \cong S^1 \times \Bbb H$. The only double cover of this, is of course, also homeomorphic to $S^1 \times \Bbb H$. Thus you conclude.
A: The homogeneous space $GL_2^+(\Bbb R)/SO(2)$ (where $GL_2^+(\Bbb R)$ is the connected component of the identity of $GL_2(\Bbb R)$) can be identified to the space of symmetric positive definite inner products $Riem$ on $\Bbb R^2$ where the map $GL_2^+(\Bbb R) \to Riem$ where $A \mapsto g_A$ we have $g_A(v,w)=g_{std}(Av,Aw)$  where $g_{std}$ is the standard inner product on $\Bbb R^2$ (it is by definition that the stabilizer of the standard inner product is $SO(2)$). Since $Riem$ is clearly convex, we have $Riem$ is contractible (and obviously paracompact and Hausdorff) hence the map $SO(2) \to GL_2^+(\Bbb R)$ is a homotopy equivalence.
To show $GL_2^+(\Bbb R)$ deformation retracts onto $SL_2(\Bbb R)$ consider the following deformation retraction
$$H(t,A)= \frac1{(1-t)+t(det(A))}A$$
A: $\newcommand{e}{\mathbf e}$
Let $\e$ be an arbitrary vector, say $(1,0)^T$. Let $f:SL(2,\mathbb R) \to \mathbb R^2 \setminus \{(0,0)\}$ be defined as $$f(M) = M\e.$$ Observe that $f$ is continuous, and never maps to the origin. Let $p: [0,1] \to SL(2, \mathbb R)$ be defined as $$p(t) = \begin{pmatrix}\cos(2\pi t) & -\sin(2\pi t) \\ \sin(2\pi t) & \cos(2\pi t) \end{pmatrix}.$$ If $p$ were contractible, then so would $f \circ p$. But $f \circ p$ is a loop in $\mathbb R^2 \setminus \{(0,0)\}$ around $(0,0)$ and hence not contractible. $\blacksquare$
