# Why is the base manifold of a Lie groupoid required to be second-countable?

I wonder why one requires that the base manifold of a Lie groupoid is second-countable?

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Isn't second-countability in the definition of manifold? en.wikipedia.org/wiki/Differentiable_manifold#Definition –  Zev Chonoles May 15 '11 at 20:41
Usually manifolds tout court are required to be second countable. –  Mariano Suárez-Alvarez May 15 '11 at 20:42
Sorry, I should state this question more carefully. Of course, Zev Chonoles and Mariano Suarez-Alvarez are right: the usual definition of a manifold requires second-countability and Hausdorff and locally euclidean. My question should merely be: At which point in the theory of Lie groupoids does one really need that the base manifold is second-countable? When constructing a Lie groupoid from a foliation one actually has to be a bit careful at this point. If one takes uncountably many charts the base manifold of the Lie groupoid won't be second-countable. –  Dave Lewis May 15 '11 at 20:57
@Dave Lewis: Can I request that you edit your question to include your comments above? (Mark the edit as an edit, so that @Zev and @Mariano 's comments still make sense.) It sounds like you have a more specific direction that you're thinking about, and in any case clearly recognize that "When constructing a Lie groupoid from a foliation one actually has to be a bit careful at this point", for example. I do not know of a good reason to have questions on MO that are only one sentence long, and there are many good reasons for including a few paragraphs. –  Theo Johnson-Freyd May 15 '11 at 22:50
@Zev, Mariano, and Dave: If you require manifolds to be second countable, then a disjoint union of manifolds is not always a manifold. Replacing second countability by paracompactness allows you to keep all good properties of second countable manifolds and makes the category of manifolds closed under coproducts, which seems like a good property to have. –  Dmitri Pavlov May 16 '11 at 4:19
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@Theo: I take my favourite non-second countable manifold: the long line $L$, and I look at the cover consisting of all of its bounded connected open subsets. The corresponding Cech groupoid is Morita equivalent to $L$. There is an obvious projection from the geometric realization of the Cech groupoid back to $L$. But there is no section of that map: that's because the cover does not admit partitions of unity. More generally, you can show that the projection does not admit a homotopy inverse. –  André Henriques May 15 '11 at 23:12