It is well-known that the universal cover $\tilde G$ of a connected Lie group $G$ has a Lie group structure such that the covering projection $\tilde G\to G$ is a Lie group morphism. Of course $\tilde G$ might not be linear even though $G$ is, but this is not the point here.


My question is: assume that $G$ is a not necessarily connected Lie group. Does there exist a Lie group $\tilde G$ and an onto Lie group morphism $\tilde G\to G$ whose restriction to the identity component of $\tilde G$ is the universal cover of the identity component of $G$? 


I assume that the answer is "no" in general, but I could not find any counter-example. 




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@Jim: Of course, the terminology "universal cover" would have been inappropriate even though such a cover existed (which as you and André pointed out, is not the case). 


I came to this question from some other direction. Namely, the $PSL\_2(R)$ action on $RP^1$ lifts to a $\widetilde{PSL\_2(R)}$ action on the universal cover of $RP^1$, and this action extends to an action of a two-sheeted cover of $PGL\_2(R)$. It is tempting to denote this cover by $\widetilde{PGL\_2(R)}$, and I wondered whether such a construction was standard. 



Is there a good definition of the universal cover for non-connected Lie groups?
The group $Pin\_-(2)$ is an example of what you're looking for.  

It can be described explicitly as a subgroup of the group of unit quaternions:  

$Pin\_-(2)=$ { $a+bi| a^2+b^2=1$ } $\cup$ { $cj+dk| c^2+d^2=1$ } $\subset \mathbb H^\times$.


Its main interesting properties are:  

- The conjugation action of $\pi\_0$ on its Lie algebra is non-trivial.  

- All the elements of the non-identity component have a non-trivial square.


There is no Lie group that is diffeomorphic to $\mathbb Z/2\times \mathbb R$
and that shares those properties.


