Last time we saw that compactly-supported de Rham cohomology is nonvanishing in the top degree for . I say that this is true for any oriented, connected -manifold . Specifically, if , then the integral of over is zero if and only if for some . That the second statement implies the first should be obvious.
To go the other way takes more work, but it’s really nothing much new. Firstly, if is supported in some connected, parameterizable open subset then we can pull back along any parameterization and use the result from last time.
Next, we again shift from our original assertion to an equivalent one: and have the same integral over , if and only if their difference is exact. And again the only question is about proving the “only if” part. A partition of unity argument tells us that we only really need to consider the case where is supported in a connected, parameterizable open set ; if the integrals are zero we’re already done by using our previous step above, so we assume both integrals are equal to . Dividing by we may assume that each integral is .
Now, if is any base-point then we can get from it to any other point by a sequence of connected, parameterizable open subsets . The proof is basically the same as for the similar assertion about getting from one point to anther by rectangles from last time. We pick some such sequence taking us from to , and just like last time we pick a sequence of forms supported in . Again, the differences between and , between and , and between and are all exact, and so their sum — the difference between and is exact as well.
And so we conclude that the map given by integration is onto, and its kernel is the image of under the exterior derivative. Thus, , just as for .