3 corrected spelling

I discover this question now, I would ad add some words to Konrad's answer. Actually there is a spectral sequence (as usual, in these cases, that means almost nothing) between the 2 cohomologies for diffeological spaces. I even had a preprint never published on that question (never published because there was no answer inside :-\ ) What I know is how to interpret the first obstruction (§ 8.30 of the manuscript): the cokernel of the first De Rham homomorphism is a class of $({\bf R},+)$ principal bundles over the space. For manifolds its gives no obstruction since every principal bundle with contractible fiber is trivial (which is a proof of the De Rham isomorphism in degree 1). For general diffeological spaces the set of such fiber bundles is some cohomology group, it's an invariant of the diffeological space (something like Picard group). I don't know much (to say nothing) about the other obstructions, I can say however something about the degree 2. Whatever... it would be very very interesting to understand geometrically these obstructions, and get from a geometrical point of view why/when, for manifolds (in particular), the De Rham homomorphism is an isomorphism. Actually it's a question I'm very interested in.

2 corrected spelling

I discover this question now, I would ad some words to Konrad's commentanswer. Actually there is a spectral sequence (as usual, in these cases, that means almost nothing) between the 2 cohomologies for diffeological spaces. I even had a preprint never published on that question (never published because there was no answer inside :-\ ) What I know is how to interpret the first obstruction (§ 8.30 of the manuscript): the cokernel of the first De Rham homomorphism is a class of $({\bf R},+)$ principal bundles over the space. For manifolds its gives no obstruction since every principal bundle with contractible fiber is trivial (which is a proof of the De Rham isomorphism in degree 1). For general diffeological spaces the set of such fiber bundles is some cohomology group, it's an invariant of the diffeological space (something like Picard group). I don't know much (to say nothing) about the other obstructions, I can say however something about the degree 2. Whatever... it would be very very interesting to understand geometrically these obstructions, and get from a geometrical point of view why/when, for manifolds (in particular), the De Rham homomorphism is an isomorphism. Actually it's a question I'm very interested in.

1

I discover this question now, I would ad some words to Konrad's comment. Actually there is a spectral sequence (as usual, in these cases, that means almost nothing) between the 2 cohomologies for diffeological spaces. I even had a preprint never published on that question (never published because there was no answer inside :-\ ) What I know is how to interpret the first obstruction (§ 8.30 of the manuscript): the cokernel of the first De Rham homomorphism is a class of $({\bf R},+)$ principal bundles over the space. For manifolds its gives no obstruction since every principal bundle with contractible fiber is trivial (which is a proof of the De Rham isomorphism in degree 1). For general diffeological spaces the set of such fiber bundles is some cohomology group, it's an invariant of the diffeological space (something like Picard group). I don't know much (to say nothing) about the other obstructions, I can say however something about the degree 2. Whatever... it would be very very interesting to understand geometrically these obstructions, and get from a geometrical point of view why/when, for manifolds (in particular), the De Rham homomorphism is an isomorphism. Actually it's a question I'm very interested in.