Why does $PSL(2,\mathbb C)\cong PGL(2,\mathbb C)$ but $PSL(2,\mathbb R) \not\cong PGL(2,\mathbb R)$? Why does $PSL(2,\mathbb C)\cong PGL(2,\mathbb C)$ but $PSL(2,\mathbb R) \not\cong PGL(2,\mathbb R)$?

You have surjective morphisms $xL(n,K)\to PxL(n,K)$ (whose kernel consists of the multiples of the identity) for $x\in\{G,S\}$, $n\in\mathbb N$ and and $K\in\{\mathbb C,\mathbb R\}$. You also have embeddings $SL(n,K)\to GL(n,K)$. Since the kernel of the composed morphism $SL(n,K)\to GL(n,K)\to PGL(n,K)$ contains (and in fact coincides with) the kernel of the morphism $SL(n,K)\to PSL(n,K)$ (it is the (finite) set of multiples of the identity in $SL(n,K)$), one may pass to the quotient to obtain a morphism $PSL(n,K)\to PGL(n,K)$, which is injective (because of "coincides with" above). The question is whether this morphism is surjective.
The question amounts to the following: given and $g\in GL(n,K)$, does its image in $PGL(n,K)$ coincide with the image of some $g'\in SL(n,K)\subset GL(n,K)$? For that to happen, there should be a $\lambda\in K^\times$ such that $\lambda g\in SL(n,K)$, and since $\det(\lambda g)=\lambda^n\det g$, one is led to search for solutions $\lambda$ of $\lambda^n=(\det g)^{-1}$, where $(\det g)^{-1}$ could be any element of $K^\times$. It is easy to see how the solvability of this polynomial equation depends on $n$ and $K$.
