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I asked a similar question on math.stackexchange but did not get any responses, so I thought I'd kick it up to mathoverflow.

In Crainic and Fernandes's "Integrability of Lie Brackets" (and the accompanying lecture notes), they use Lie II to prove the equivalence between $A$-paths (Lie algebroid morphisms $TI \to A$, where $TI$ is the tangent bundle over the unit interval) and $G$-paths (functors from the pair groupoid over $I$ into $G$). This is, of course, rock solid as a proof, but I find it a bit unsatisfying to have a fairly important step of a proof handled by applying a very powerful theorem to one of the simplest possible cases where it applies.

I first tried unwinding one of the original proofs of Lie II for groupoids from Moerdijk and Mrcun's book, where I assumed the source-simply-connected Lie groupoid in question is the pair groupoid over $I$. However, I once again found myself using powerful theorems about foliations while dealing with one of the simplest possible cases (the foliation of $I \times I$ by $I$). This feels like a result that should be amenable to elementary techniques and is nestled away in a masters thesis somewhere or a paper from the early days of Lie algebroids, but I can't seem to find a reference. I'm also sure that someone working in this field can probably pull the construction off the top of their head, but I'm having trouble finding the construction myself.

In summary:

  1. Does anyone know where I can find an explicit/direct proof of the bijection between functors on the pair groupoid over the unit inverval $I$ into a Lie groupoid $G$ and Lie algebroid morphisms $TI \to Lie(G)$?
  2. It seems as though the construction in (1) should be derivable from something like "the pair groupoid over a vector space $V$ integrates the trivial lie algebroid $TV \cong V \times V \xrightarrow{\pi_0} V$". This would be the ideal construction to find.

Update 1: The natural next step is to consider the path integration method on a Lie group. There is an explicit proof of the bijection in Proposition 1.13.4 of Duistermaat and Kolk's Lie Groups.

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I suspect you've probably figured it out by now but here is an answer anyway:

  1. Here is a sketch of the most low-tech way of accomplishing the task.

First, there is a bijection between groupoid homomorphisms $\Gamma \colon I \times I \to G$ and smooth functions $\gamma \colon I \to G$ such that $\gamma(0)$ is a unit and $\gamma$ is tangent to the source distribution $\ker d s$. This bijection is as follows: $$ \Gamma(t_1, t_0) := \gamma(t_1) \gamma(t_0)^{-1} \qquad \gamma(t) := \Gamma(t,0) $$

Now if you are given a path $\gamma \colon [0,1] \to G$ which is tangent to the source fibers and such that $\gamma(0)$ is a unit, then one can construct a function $a \colon [0,1] \to A$ by: $$ a(t) = d L_{\gamma(t)}^{-1} \gamma'(t) $$ where $L_g \colon s^{-1}(t(g)) \to s^{-1}(s(g))$ is the left translation map.

Perhaps the trickiest part is to show the correspondence $\gamma \to a$ ends up being a bijection. Surjectivity is a consequence of existence of solutions to ODEs and infectivity is a combination of the assumption that $G$ is source-simply-connected and uniqueness of solutions to ODEs.

  1. the result you are asking for is not proved but appears in Lectures on Integrability of Lie Brackets by Crainic and Fernandes as exercise 30. The exercise is not difficult:

The source map of the pair groupoid $M \times M$ over $M$ is just projection to the second component. The identity section is the diagonal map $\Delta \colon M \to M \times M$ and there is an obvious identification of $TM$ with $A := \ker d \pi_2 |_{\Delta(M)}$. The target map is projection to the first component which means that the anchor map $\rho \colon A \to TM$ is a vector bundle isomorphism. Since $\rho$ preserves the Lie bracket of both algebroids it is an algebroid isomorphism.

To show that the Lie algebroid of the fundamental groupoid of $M$ is $TM$ you can accomplish this by observing that $(pi_1 \times \pi_2) \colon \Pi_1(M) \to M \times M$ is a groupoid homomorphism and a local diffeomorphism.

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