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).

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.

- 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.
- 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.

somethingis a surjective submersion, I can't recall right now if it's that, or if it's just an iso. $\endgroup$ – David Roberts Sep 8 at 23:21