Here is one answer, but also sort of a no-go observation.

We are interested in creating a bicategory whose objects are Lie algebroids. The 1-morphisms and 2-morphisms are something yet to be determined. We have a couple simple requirements that we are going to demand:

- We want a functor from the bicategory of Lie groupoids and bibundles to this bicategory.
- This functor should take a Lie groupoid to its underlying Lie algebroid.
- The naive notion of morphism of Lie algebroid gives rise to a 1-morphism in this bicateogry, and isomorphism become equivalences.

Now I will describe a bicategory which minimally satisfies these conditions, but I don't think that this bicategory of Lie algebroids is going to be very interesting. The reason is that the above requirements force you to identify too many things.

For example take any space X and any cover U of the space. We can form two groupoids out of this which are Morita equivalent. These are X, viewed as a groupoid with just identities and the Cech groupoid $U \times_X U \rightrightarrows U$. These are equivalent groupoids in the bibundle bicategory and so must be sent to equivalent Lie algebroids. (In a certain sense the bibundle bicategory is what you get when you take Lie groupoids, functors, etc. and force these two types of groupoids to be equivalent. More on this below.)

However their Lie algebroids are very simple and appear very different. They are just the trivial Lie algebroids over X and U, respectively. In particular this second one has no information about the fiber products $$U \times_X U.$$ This means that the tangent bundle of X and the tangent bundle of U must be equivalent objects in this hypothetical bicategory of Lie algebroids, whenever U covers X.

Following through with similar examples allows us to see that any time we pull a Lie algebroid back by a cover of its base we get an "equivalent" Lie algebroid. In particular this means that every Lie algebroid will be equivalent to one on a trivial bundle.

Maybe I am wrong and this is still an interesting bicategory, but it feels like we're loosing too much information. In any event this suggests what your hypothetical bicategory of Lie algebroids actually looks like.

Let's set up some terminology first. Suppose I have a Lie algebroid with base space X. Suppose further that I have a surjective submersion of the base $U \to X$. Then I can pull-back the Lie algebroid on X to one on U. We will call the morphism of Lie algebroids from U to X a *weak equivalence".

So the bicategory you are after should have object Lie algebroids and the morphisms should be spans of Lie algebroids $$X \stackrel{\sim}{\leftarrow} U \rightarrow Y$$ where $U \to X$ is a weak-equivalence.

The 2-morphisms are not just naive morphisms of spans. They are more complicated. This is why your previous MO question doesn't apply. What we are doing is inverting the weak equivalences to make them, well, equivalences. We want to do it in such a way that we still have a bicategory and that it has the obvious universal property (functors out of it are the same as weak equivalence inverting functors out of Lie algebroids). This is sometimes called the derived localization.

Fortunately there is some systematic machinery to accomplish this, devoloped by Dorette Pronk. Some of it is described at the nlab, but the full story is in her paper "Etendues and stacks as bicategories of fractions". Among other things, this paper contains a description of the bicategory of Lie groupoids and bibundles as an example of this sort of derived localization, and so that is a good example to compare with.

From this description it is also clear that the derived localization of Lie algebroids along the weak equivalences is going to be the universal (initial) thing which satisfies the three properties outlined above.

The question remains though: is this an interesting bicategory? I don't have an answer for this. Perhaps you have a use for it?

and smooth natural transformations. But I don't see how to define "natural transformations of morphisms of Lie algebroids". So the only definition of "spans of Lie algebroids" I can write down is boring. – Theo Johnson-Freyd Jul 31 '10 at 2:48