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Definition I am aware of for Lie groupoid is that (among other things) the source and target maps $s,t:\mathcal{G}_1\rightarrow \mathcal{G}_0$ are submersions.

On page 9 of Du Li's thesis Higher Groupoid Actions, Bibundles, and Differentiation (arXiv:1512.04209) the author ask $s,t:\mathcal{G}_1\rightarrow \mathcal{G}_0$ to be surjective submersions. This is the only place where I saw this kind of requirement.

What properties do we get easily assuming that source, target maps are surjective submersions rather than just submersions.

Lie groupoid $(M\rightrightarrows M)$ coming from Manifold $M$ is such that source, target maps are surjective submersions.

Lie groupoid $(G\rightrightarrows *)$ coming from a Lie group $G$ is such that source, target maps are surjective submersions.

Are there natural examples of Lie groupoids where source/target maps are not surjective submersions?

This condition continue on all results. For example, given a morphism of Lie groupids $\mathcal{G}\rightarrow\mathcal{H}$ and $\mathcal{K}\rightarrow \mathcal{H}$ the pullback $\mathcal{G}\times_{\mathcal{H}}\mathcal{K}$ is a Lie groupoid if $t\circ pr_2: \mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1\rightarrow \mathcal{H}_0$ is a sujective submersion.

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    $\begingroup$ If $s$ and $t$ are not surjective then $\mathcal G$ is not a category (and in particular not a groupoid), because every object in a category has an identity morphism. $\endgroup$
    – mme
    Commented Feb 12, 2019 at 23:05
  • $\begingroup$ @MikeMiller Ofcourse they are surjective... Given $a\in \mathcal{G}_0$ I can just take $1_a:a\rightarrow a$ in $\mathcal{G}_0$ that says both $s,t$ are surjective... It was asked 5 hours ago i.e., 4 am for me.. I should have just slept off.. My mind did not work and I did not realise that then... I now feel like a real stupid :D I was also rechecking the definition in arxiv.org/pdf/math/0203100.pdf and it says it is just submersion :D :D It is already surjective so they just did not mention :P :P $\endgroup$ Commented Feb 13, 2019 at 3:32

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Edit I just realised your misconception: the source and target maps are automatically surjective since they both have a section, namely the unit map. So asking that they are surjective submersions or just submersions are equivalent.


This is the only place where I saw this kind of requirement.

It's been literally the definition since the 1980s.

If the source and target map are not assumed to be (surjective) submersions then you need to build into your definition the requirement that for a Lie groupoid $X_1 \rightrightarrows X_0$ the pullback $X_1\times_{s,X_0,t} X_1$ is a manifold. Charles Ehresmann for instance took this point of view in the 1950s. But then, many other constructions break without assuming source/target are submersions as you need to pull these back all the time. Submersions are also the class of maps of manifolds that are the 'saturation' of the singleton Grothendieck pretopology consisting of coprojections $\coprod_\alpha U_\alpha \to M$ of open covers, and the whole point of Lie groupoids is that they present differentiable stacks on the site of manifolds and open covers. This is not something that was considered before the late 1980s/1990s. The paper Groupoids in categories with pretopology by Meyer and Zhu in TAC gives the general theory of groupoids internal to categories with a class of maps like that of submersions. It is a fluke of the category of sets that (split) surjections have all the required properties and so no further constraints are needed. A similar thing happens in the world of Deligne–Mumford stacks, where one assumes groupoids in schemes have étale source and target, or Artin stacks, where the source and target are required to be smooth maps of schemes.

Are there natural examples of Lie groupoids where source/target maps are not surjective submersions?

Not really, because they are useless. You can come up with some kind of topological groupoid where the object and arrows spaces admit manifold structures, and the source, target etc maps are smooth, but... why?

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  • $\begingroup$ Ofcourse they are surjective... Given $a\in \mathcal{G}_0$ I can just take $1_a:a\rightarrow a$ in $\mathcal{G}_0$ that says both $s,t$ are surjective... It was asked 5 hours ago i.e., 4 am for me.. I should have just slept off.. My mind did not work and I did not realise that then... I now feel like real stupid :D Sorry for wasting your time.. I feel guilty.... $\endgroup$ Commented Feb 13, 2019 at 3:27
  • $\begingroup$ I was also rechecking the definition in arxiv.org/pdf/math/0203100.pdf and it says it is just submersion :D :D It is already surjective so they just did not mention :P :P $\endgroup$ Commented Feb 13, 2019 at 3:31
  • $\begingroup$ This does not deserve to be a question on MO or question on anywhere... it is not a misconception it is just because of my sleepy feeling... of course I do know that source and target maps are surjective,,, what should I do now?? Should I just delete it ? $\endgroup$ Commented Feb 13, 2019 at 3:42
  • $\begingroup$ Well, I answered a different question, namely why should they be submersions, and that information is worth leaving there. $\endgroup$
    – David Roberts
    Commented Feb 13, 2019 at 4:26
  • $\begingroup$ Yes, I see why they are submersion,, first reason I realise why $s,t$ is submersion is to make sure $\mathcal{G}_1\times_{\mathcal{G}_0}\mathcal{G}_1$ is a manifold (pullback is a manifold if one of the maps is submersions) so that you can talk about multiplication/composition map being smooth... after this also many times I used source/target map are submersions so that pullback is manifold... I do not know much about Deligne Mumford stacks and all that.. may be some day I will know and this answer might be useful... thanks :) $\endgroup$ Commented Feb 13, 2019 at 4:59

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