I was woundering, is $PSL_2(\mathbb R)$ a lie subgroup of $SL_2(\mathbb R)$?
I don't quite know how to check if the inclusion is an immersion because I don't know how to work with the tangent space at points $p\not=e$ (at $e$ this should be trivial, because it's simply the lie algebras, which are equal for both groups).
If the answer is yes, then $\exp:\:\mathfrak{sl_2}(\mathbb R) \to PSL_2(\mathbb R)$ should be just the exponentiation of $SL_2$ composed with the projection $SL\to PSL$, right? But if the answer is no, what's the exponentiation map?
As I suggested above, if $H$ is a normal subgroup of $G$, then it's a standard well-definedness check that there is a group homomorphism $\beta$ injecting $G/H$ into $G$ (with $\pi\circ\beta$ the identity, $\pi\colon G\to G/H$ being the obvious map) if and only if there is a group homomorphism from $G$ to $H$ that is the identity on $H$. So, in our case, we ask if there is a group homomorphism from $G=SL_2(\Bbb R)$ to $H=\{\pm I\}$ that is the identity on $H$. To see that the answer is no, just note that $-I$ is a square in $G$.