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In the definition of a Lie groupoid, the source and the target maps are required to be submersions. I want to know the reason for that. I write down definitions below. See also https://en.wikipedia.org/wiki/Lie_groupoid.

A groupoid is a (small) category in which every morphism is an isomorphism. A Lie groupoid is a groupoid with additional data satisfying the followings.

(1) the set of objects is a smooth manifold $M$.

(2) the set of morphisms is a smooth manifold $\mathcal{G}$.

(3) The multiplication, inverse, and the identity assigning maps are smooth.

(4) The source and the target maps $s,t\colon \mathcal{G} \rightarrow M$ are submersions.

Here I have a question on the 4th condition. I guess this condition is to make $\mathcal{G}$ a fiber bundle over $M$. By Ehresmann's theorem, a smooth map is a fiber bundle if it is surjective, submersive, and proper. Two maps $s,t$ are clearly surjective, but since they may not be proper we cannot apply the theorem. Then what is the advantage of assuming the 4th condition?

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Because in this way, for all objects $x,y \in M$, the subset $$\hom(x,y) = s^{-1}(x) \cap t^{-1}(y) = (s,t)^{-1}(x,y) \subset \mathcal{G}$$ is a submanifold of $\mathcal{G}$ (because $(s,t) : \mathcal{G} \to M$ is then a submersion, so all its values are regular, and the preimage of a regular value is a submanifold).

This is also useful because the category of manifolds does not have all pullbacks, and when you define composition you need something like $$\circ : \{(g,f) \in \mathcal{G}^2 : s(g) = t(f) \} = \mathcal{G} \times_{(s,t)} \mathcal{G} \to \mathcal{G}.$$ In this way, the extra condition ensures that this last set is a manifold too, so that you can do everything in the realm of manifolds.

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  • $\begingroup$ You can find more info about this in the nLab article on Lie groupoids. $\endgroup$ – Najib Idrissi Jul 13 '15 at 9:22
  • $\begingroup$ the proof that $hom(x,y)$ is a manifold is not correct, because the map $(s,t)$ goes from $\mathcal{G}$ to $M\times M$, so it is not a submersion in general. $\endgroup$ – Studzinski May 20 '16 at 20:40
  • $\begingroup$ Sorry to wake up an old answer. As said by @Studzinski your proof of $hom(x,y)$ being a submanifold does not seem to be correct. can you say something about your justifiction. $\endgroup$ – Praphulla Koushik Jun 1 '18 at 17:18

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