Let $G$ be a group locally isomorphic to $SL(2,\mathbb R)$, and with finite center, then $G$ is a covering space of $PSL(2,\mathbb R)$ I'm reading from a book about lie groups and representations.
The author mentions without prove, what should be trivial, that:
If $G$ is a group locally isomorphic to $SL(2,\mathbb R)$ and $G$ has finite center, then there is a finite sheet covering $f:\:G\to PSL(2,\mathbb R)$ such that $Z_G=\ker f$.
I can't see why this is true. My attempt was trying to use the adjoint group $Ad(G)$ and its relation to the inner automorphism group (which for $SL(2,\mathbb R)$ I know it to be $PSL(2,\mathbb R)$ , to show that the adjoint representation would satisfy the conditions.
 A: I think your idea works. I guess we can assume that the finite center Lie group $G$ has an isomorphism of its Lie algebra 
$\phi :\frak{g} \to \frak{sl}_2(\mathbb{R})$.
This induces isomorphisms ($\phi^*(M) = M\circ \phi$) of Lie algebras in the top row and an isomorphism of Lie groups in the bottom row of
$\require{AMScd}$
$$
\begin{CD}
\text{Der}(\frak{sl_2}) @>>> \operatorname{End}(\frak{sl}_2) @>{\phi^*}>> \operatorname{End}(\frak{g}) @<<< \text{Der}(\frak{g})
  \\
& @V{\text{exp}}VV @V{\text{exp}}VV  \\ 
 (\text{Aut}(\frak{sl_2)})^\circ  @>>> \operatorname{GL}(\frak{sl}_2) @>{\phi^*}>> \text{GL}(\frak{g}) @<<< (\text{Aut}(\frak{g)})^\circ
\end{CD}$$
Note, for semisimple $\frak{g}$, $\text{ad}(\frak{g})$ coincides with the derivations and the corresponding Lie group is the connected component of the automorphisms. See for example the relevant section in Knapp's book on Lie groups.
From the diagram we can conclude that $(\text{Aut}(\frak{sl_2)})^\circ$ is isomorphic to $(\text{Aut}(\frak{g)})^\circ$ and each one is the image of $\text{Ad}$ under $\text{SL}_2(\mathbb{R})$ and $G$ respectively.
Thus $G$ covers $\text{Ad}(G) \approx \text{PSL}_2(\mathbb{R})$ with the kernel being the center.
