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?


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.

  • $\begingroup$ 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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