# Identifying Lie groupoids among smooth groupoids

I have been approaching groupoids in the category of smooth manifolds using methods from essentially algebraic theories/limit sketches. Are there any results that identify Lie groupoids amongst internal groupoids in the category of smooth manifolds (without the advance knowledge that $$s,t: G \to M$$ are submersions).

• Are you defining, as part of the data, additional to $X_1$ and $X_0$ and source/target etc, a manifold $X_2$ and three maps $X_2 \to X_1$ (two "projections" and "composition") so that $X_2 \simeq X_1\times_{X_0} X_1$? It's not clear to me how one extracts from the existence of a pullback square of manifolds that both of the maps in the cospan part are submersions. – David Roberts Sep 8 at 22:59
• In practice, I have a (symmetric) simplicial manifold that satisfies some version of the Segal conditions. I should edit my question - I guess where I'm confused is whether the conditions for "geometric groupoids" in papers by Henriques or Getzler and Behrend will force a simplicial manifold to be a Lie groupoid. – Ben MacAdam Sep 8 at 23:12
• Oh, ok. I was assuming you were working from the older Eheresmann-style sketch, where ambient cats without finite limits were the norm, so the required limits and their cones were built in. Since you can say an internal groupoid is an appropriately coskeletal simplicial object, you can truncate just above that, and only deal with a finite diagram (I almost wrote "Segal condition" but decided to write out the only pertinent iso!). But saying what "coskeletal" means might require saying something is a surjective submersion, I can't recall right now if it's that, or if it's just an iso. – David Roberts Sep 8 at 23:21
• Well I'm using a Kelly-style enriched sketch so I'm able to cheat a bit and force the nerve of the groupoid to exist, and further require that all of the pullbacks be preserved by the tangent functor. – Ben MacAdam Sep 8 at 23:30

Well this is embarrassing, I asked this question after spending a weekend thinking about it staring at the definition all day and two hours later I have a partial answer - it seems that this is equivalent to requiring the tangent projection $$p: T\mathcal{G} \to \mathcal{G}$$ be a bifibration.
1. Recall from the theory of Grothendieck fibrations and opfibrations, for any functor $$q: \mathbb{C} \to \mathbb{D}$$, if $$u:q^{-1}(A) \to q^{-1}(B)$$ over $$f: A \to B$$ is an isomorphism, then $$u$$ is both cartesian and cocartesian. For a functor between groupoids, any map is (co)cartesian above its image.
2. A morphism $$f: M \to N$$ of smooth manifolds is a submersion if the naturality square with the tangent projection $$p: T \Rightarrow id$$ is a weak pullback. So for any $$(a,b):X \to M \times TN$$ where $$f(a) = p(b)$$, there exists a map $$c: X \to TM$$ so that $$p(c) = a, Tf (c) = b$$.
Now, recall that our smooth groupoid $$\mathcal{G}$$ has all of its pullbacks preserved by the tangent functor. This means that we have a functor $$p: T \mathcal{G} \to \mathcal{G}$$. The condition that target be a submersion is equivalent to asking for any map $$a \xrightarrow{w} b$$ and tangent vector $$\gamma:TM$$ so that $$p(\gamma) = b$$, there is a map $$\gamma' \xrightarrow{\omega} \gamma$$ above $$w$$. Because $$\omega$$ is an isomorphism, it will be a cartesian map, so this is equivalent to requiring that $$p$$ be a fibration. A similar line of reasoning for the source map leads to the conclusion that $$p$$ is also an opfibration. If $$s,t$$ are both submersions, then $$p$$ is a bifibration.