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Aug 29, 2021 at 22:29 vote accept V. Rogov
Aug 29, 2021 at 21:45 comment added Will Sawin I guess the recentlyd-defined condensed group would be a similar strategy, also involving enlarging the category of topological groups to a category of sheaves of groups on a site.
Aug 29, 2021 at 15:04 comment added Dmitri Pavlov @V.Rogov: The object $H^i(X,F)$ is a smooth group, i.e., a sheaf of abelian groups on the site of smooth manifolds. If this sheaf is representable, then we can identify $H^i(X,F)$ with a Lie group. Otherwise $H^i(X,F)$ is a smooth group that is not a Lie group. So one point of my answer is that the category of Lie groups must be enlarged (following Grothendieck's advice) to a good category (smooth groups in this case).
Aug 29, 2021 at 10:23 comment added V. Rogov I didn't understand you answer completely yet, but are you claiming that if $\mathcal{F}$ is a sheaf of Lie groups, then $H^i(X, \mathcal{F})$ are Lie groups? This confuses me a lot, say I don't understand why $H^i(X, \mathcal{F})$ should be Hausdorff.
Aug 29, 2021 at 2:54 history edited Dmitri Pavlov CC BY-SA 4.0
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Aug 29, 2021 at 2:46 history answered Dmitri Pavlov CC BY-SA 4.0