Recall the related notions of [Lie groupoid](http://ncatlab.org/nlab/show/Lie+groupoid), [Lie algebroid](http://ncatlab.org/nlab/show/Lie+algebroid), [generalized morphism of Lie groupoids](http://ncatlab.org/nlab/show/Lie+groupoid#2CatOfGrpds), and [cohomology of Lie algebroid](http://ncatlab.org/nlab/show/Lie+algebroid#the_cealgebra_of_a_vector_bundle_with_anchor_19). Henceforth, I will drop the word "Lie" for all those things listed above, because I want to reuse it: there is a functor "Lie" from the 1-category of groupoids to the 1-category of algebroids. There is also a (contravariant) functor "cohomology" from the 1-category of algebroids to the 1-category of graded commutative algebras. However, their composition does not extend to generalized morphisms (which are really 1-morphisms in a 2-category of groupoids).


In particular, the algebroids for equivalent groupoids need not have isomorphic cohomology. A good example is as follows. For any manifold $M$, there is a "pair" groupoid $M\times M \rightrightarrows M$ with object the points in $M$ and a unique morphism between each pair of points. In fact, this groupoid is equivalent to the groupoid $\{\text{pt}\}$ with one object and one morphisms. But $\operatorname{Lie}(M\times M \rightrightarrows M) = {\rm T}M$ is the tangent groupoid, and the cohomology of this algebroid is the de Rham cohomology of $M$, which need not be trivial.


My question is: 



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> If $G\_1,G\_2$ are two equivalent groupoids, and if both are source-simply-connected, does it follow that the cohomologies of the algebroids $\operatorname{Lie}(G\_1)$ and $\operatorname{Lie}(G\_2)$ are isomorphic?
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There is a converse question, for which I am less optimistic, and that I haven't thought much about myself:



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> If $G\_1,G\_2$ are source-simply-connected groupoids and the cohomologies of $\operatorname{Lie}(G\_1)$, $\operatorname{Lie}(G\_2)$ are isomorphic, does it follow that $G\_1,G\_2$ are equivalent?
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My motivation is the following. There ought to be (but there is not, although almost) a "Lie III theorem" that says that the categories of algebroids and of source-simply-connected groupoids are equivalent. Groupoids present stacks, and the question becomes what "stack-like" thing algebroids present. If the answers to both questions are "yes", then the "stack-like thing" presented by an algebroid just *is* its cohomology. But probably the answers are not both "yes" — even answers "yes, no" means that, well, the cohomology doesn't entirely determine the stack, but it is an invariant.



Is the cohomology of the corresponding Lie algebroid an invariant under equivalence of source-simply-connected Lie groupoids?
I may be misinterpreting what you said (in particular, I don't know what you mean by "source simply connected"), but it sounds like you basically answered your own question in the negative: By employing the pair construction $M \times M \rightrightarrows M$, it suffices to find two simply connected manifolds with nonisomorphic cohomology. The standard examples are the point and the sphere $S^2$.


