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A fairly reasonable interpretation of Lie II and Lie III seems to be that the category of Lie algebras is a coreflective subcategory of the category of Lie groups, so that the Lie group integrating a Lie algebra satisfies a couniversal property. While this fails for Lie groupoids and algebroids, Crainic and Fernandes's result tells you exactly when a groupoid satisfying this couniversal property exists.

I'm a bit confused about the move to categories of infinite dimensional manifolds, such as Banach or convenient manifolds, or settings like smooth sets. At first glance, the derivative of a group(oid) operation will be preserved by the inclusion of smooth manifolds into one of these categories since they all seem to preserve transverse limits. However, at first glance it seems unlikely that the couniversality of the integrating group(oid) would be preserved by this inclusion, and there could be a different group(oid) satisfying the couniversal property in the new category. Is this problematic, or just a natural consequence of working in a more general category?

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    $\begingroup$ Probably not helpful, but certainly in the more exotic infinite-dimensional setting, the exponential map may fail to be sufficiently well-behaved. I note you didn't mention Fréchet manfiolds or worse, so probably you are trying sidestep that issue? $\endgroup$
    – David Roberts ♦
    Commented Oct 30, 2020 at 0:00
  • $\begingroup$ Yeah, I guess I'm trying to keep things in the realm of universal algebra. Without the exponential map, you don't have the coreflection so I would say Lie's theorem fails. $\endgroup$
    – Ben MacAdam
    Commented Oct 30, 2020 at 23:04
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    $\begingroup$ I think one usually is careful defining Lie groups in that context, so that the Lie algebra is the model space for the underlying manifold of the Lie group, and the exponential map does what it should. But you are right in that this is already an interesting question before that is necessary! $\endgroup$
    – David Roberts ♦
    Commented Oct 30, 2020 at 23:11
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    $\begingroup$ I just found that Lie II holds for Banach Lie groups, (see eg Theorem 8 in Miguel M. R. Moreira, On Lie's third theorem people.math.ethz.ch/~mimoreira/lie_3rd_thm.pdf) so that's a partial answer, at least. But Lie III can fail for Banach Lie algebras, with an example in those notes. $\endgroup$
    – David Roberts ♦
    Commented Nov 1, 2023 at 6:35

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