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It is a known theorem that an internal equivalence of Lie groupoids (finite dimensional manifolds!) - that is an equivalence in the 2-category of Lie groupoids, smooth functors and transformations - is a weak equivalence: a fully faithful essentially surjective functor. Here we say a functor is essentially surjective is the map expressing this fact is not only surjective but a surjective submersion.

I'm interesting an analogues of this fact for Frechet Lie groupoids - groupoids internal to the category of Frechet manifolds, where the source and target are submersions of Frechet manifolds (this is stronger than surjective on tangent spaces - need local charts where the map looks like projection out of a direct sum).

The proof for Lie groupoids relies on the fact that Lie groupoids admit local bisections through every arrow $g$. These are maps $X_0 \supset U \stackrel{f}{\to} X_1$ where $s(g) \in U$, an open subset of $X_0$ such that $f(s(g)) = g$, and $t\circ f:U \to X_0$ is an open embedding. So far so good, the existence of local bisections depends on the characterisation of a submersion as locally a projection out of a direct sum, but with a small twist, which I haven't thought about, but don't expect to cause trouble.

The problem is showing that the 'surjective implies submersion' part of the proof, which uses a different characterisation of submersions of finite-dimensional manifolds, namely that admit local sections through every point in their codomain. This is false in the general Frechet case, but it doesn't mean the proof couldn't be rewritten to use the other characterisation of submersions (locally a projection).

My question is: has this been done?

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The part about the functor being fully faithful is no problem at all and can be shown arrow-theoretically with a little Yoneda thrown in. It is the surjective submersion part which is fragile. –  David Roberts Jan 30 '12 at 6:23
    
It seems to me that a definition of "essentially surjective" which doesn't include the case when the map in question is split epi can't possibly be right. (In particular, split epis cover in any Grothendieck topology.) Unless I misunderstand the question? –  Mike Shulman Feb 2 '12 at 18:34
    
In the case of fin. dim. Lie groupoids if the map in question is a split epi, the structure of the Lie groupoid means it is a submersion. But split epis are not submersions in general. And this definition uses a pretopology, not a Grothendieck topology. Perhaps my question could also ask that perhaps for Frechet Lie groupoids, are submersions even the right pretopology to use? They probably are, but I can't say with 100% certainty. –  David Roberts Feb 2 '12 at 23:59
    
Obviously the notion of "right" is subjective, but it seems to me that if $f\colon X\to Y$ is a smooth functor such that there is a smooth function assigning to every object $y\in Y$ an object $x\in X$ and an isomorphism $f(x)\cong y$, then why wouldn't you want to call $f$ "essentially surjective"? –  Mike Shulman Feb 5 '12 at 7:30
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Just a (sort of) layman's comment: it seems to me that you need a characterization of surjective submersions that involves the implicit function theorem. If that's really the case, I'd guess that the result still holds true in the tame Fréchet category, thanks to the Nash-Moser implicit function theorem. –  Pedro Lauridsen Ribeiro Apr 20 '13 at 3:10
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2 Answers

I don't know whether this is really an answer, but since the question is open for a while now I'll give it a try. If I am not mistaken, then the Lie groupoids are weakly equivalent if and only if the associated 2-functors $\operatorname{Man}^{op}\to \operatorname{Grp}$ are equivalent and the argument should go through also in the infinite-dimensional case as well (btw, why do you take Fréchet-Lie groupoids, I would expect that -unlike in the Banach case- the neither the metrisability nor the completeness does any job for you). That the functors are equivalent if the Lie groupoids are even strongly equivalent is obvious. Or am I missing something?

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The problem is that a map of Fréchet manifolds admitting (enough) local sections is not enough to make it a submersion: it is a strictly stronger notion. And the reason one wants submersions is to ensure certain pullbacks exist, as in the finite-dimensional case. –  David Roberts Aug 23 '13 at 0:30
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This is partly an answer (or maybe more of an extended comment) and partly a response to Christoph's answer, but too long for a comment. One needs to be careful what you mean by "associated 2-functors". The most naturally associated such $2$-functor to a (say Frechet) Lie groupoid $G$, is the functor $\tilde y \left(G\right)$ assigning a manifold $M$ the groupoid of internal functors from $M$ (viewed as a Lie groupoid) to G. The stackification of this functor is the stack of principle $G$-bundles. A morphism $\varphi:G \to H$ is a strong equivalence if and only if the induced map $$\tilde y\left(G\right) \to \tilde y\left(H\right)$$ is an equivalence. A good definition of a "weak equivalence" is if the induced morphism between their associated stacks is an equivalence. Perhaps this implies that the map defining "essential surjectivity" is a submersion of sorts, but if this is not the case, perhaps one could get away with demanding this, and end up with an equivalent saturated class of morphisms to invert. At any rate, it's still an interesting question.

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Herein lies the problem: it gets a bit circular if I start defining a weak equivalence to be a map which induces an equivalence of stacks. In any case, I can get the result I want without depending on whether the approach I outline in the question works or not. –  David Roberts Feb 9 '13 at 0:15
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Or rather, not circular, but the statement of the desired theorem is rather shallow :-) –  David Roberts Feb 9 '13 at 0:16
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