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I am wondering how to show that the Lie functor taking Lie groups to Lie algebras is faithful? In particular, I am looking for a constructive proof, since I am working in the context of synthetic differential geometry (and thus restricted to intuitionistic logic).

Sorry if this is an obvious question, but my algebra background isn't that strong, and I mostly taught myself what little Lie theory I know.

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  • $\begingroup$ This is false unless you restrict to connected Lie groups. I am not sure what's constructive, but what proofs have you seen? There's two that I know: one that first passes through formal group laws (in Serre's book on Lie groups) and one that passes through foliations (you can find this in Lee's book). $\endgroup$ – user98602 Jan 4 '16 at 14:34
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There is a reference with a proof given here. It says "the second Lie theorem establishes that this functor is fully faithful". See also here.

For non-connected Lie groups the Lie functor is not faithful, see page $35$ here for "explicit counterexamples".

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    $\begingroup$ "Faithful" is not synonymous with "injective on objects"... In fact these two properties are quite unrelated. Did I misunderstand your answer? $\endgroup$ – Najib Idrissi Jan 4 '16 at 14:38
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    $\begingroup$ Sorry, this was not correct. I included the nLab reference now. $\endgroup$ – Dietrich Burde Jan 4 '16 at 14:48
  • $\begingroup$ Thanks! For some reason I had thought that the Lie functor was faithful for the entire category of Lie groups. Do you happen to know of an explicit counterexample by the way (i.e. a pair of distinct Lie group homomorphisms which map to the same Lie algebra homomorphism under the Lie functor?) $\endgroup$ – ಠ_ಠ Jan 5 '16 at 0:16
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    $\begingroup$ Yes, there are many explicit counterexamples. Just consider any two non-connected Lie groups $G$ and $H$ such that their respective identity connected components $G_I$ and $H_I$ are Lie groups themselves - see my edit. $\endgroup$ – Dietrich Burde Jan 5 '16 at 11:10
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    $\begingroup$ To be rather silly: $G = \Bbb Z/2$, $H = *$. $\endgroup$ – user98602 Jan 6 '16 at 18:42

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