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could someone please clarify the definitions of extensions of topological groups and Lie groups.

For topological groups, what I see in most papers is as follows: An extension of topological groups $0 \to N \stackrel{i}{\rightarrow} G \stackrel{\pi}{\rightarrow} Q \to 0$ is an algebraically exact sequence of topological groups such that $i$ is closed and continuous and $\pi$ is open and continuous.

However, for Lie groups, it is defined as follows: An extension of Lie groups $0 \to N \stackrel{i}{\rightarrow} G \stackrel{\pi}{\rightarrow} Q \to 0$ is an algebraically exact sequence of Lie groups such that both $i$ and $\pi$ are smooth and $\pi$ has a local smooth section in a neighbourhood of identity in $G$.

Are these the standard definitions? Does these definitions imply that an extension of Lie groups give an extension of topological groups by restricting to the topological structure?

Kindly provide any references.

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I've never thought it necessary to have a context-independent definition of "extension" for the category or Lie groups. So perhaps I'm biased but I don't believe I've noticed a standard definition. If someone forced me to write a definition I'd probably remove your local section condition, and demand that $\pi$ be an onto function. –  Ryan Budney Jul 18 '11 at 14:00
    
@Ryan: You certainly want $Q$ to be diffeomorphic to $G/N$ via the map induced by $\pi$, thus $\pi$ had better be a submersion (and thus admit local sections). I'm not sure if that's automatic (I'd say yes, but I need to think a bit). –  t.b. Jul 18 '11 at 14:10
    
Thanks all. With the definitions that I have written, does the second question answers in an affirmative? –  beginner Jul 18 '11 at 14:53
    
The condition you have on $\pi$ in an extension of Lie groups implies that $\pi$ is a submersion (since the differential at the identity admits a right inverse), hence $\pi$ is an open map, see e.g. Cor. 2.3 on p. 11 (p.14 of the pdf) here. Clearly, $i$ must be closed and continuous. –  t.b. Jul 18 '11 at 15:25
    
Thank you very much Theo. –  beginner Jul 18 '11 at 16:41

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