Computing Rham Cohomology Suppose that we have a $C^{\infty}$ manifold $X$ with and atlas $\mathcal{A}=$($U_{\alpha},\varphi_{\alpha}$) such that for every two intersecting open sets $U,V \in \mathcal{A}$ the intersection is connected.
Is it true that when we take locally defined exact forms we can glue them together? I mean given two exact forms $df,dg$ defined on $U, V$ such that $U \cap V \neq \emptyset$ we know that $d(df-dg)=0$ in the intersection. What connectedness tell us now? Is it similar to what happens with functions?
Can this be used to show that De Rham cohomology should be zero?
 A: As @TedShifrin pointed out, just assuming connectedness of the intersections is not enough. But there is a way to compute de Rham cohomology if you have an open cover with the property that all sets in the cover, and all intersections of finitely many sets in the cover, are contractible. Such a cover is called a good cover, and there always exists such a cover. For example, you can take the sets to be convex geodesic balls with respect to some Riemannian metric. 
If there's a finite good cover, then you can compute the de Rham groups directly from the cover using the Mayer-Vietoris theorem; but the computation isn't as straightforward as checking connectedness.  This is described very nicely in Bott and Tu's book Differential Forms in Algebraic Topology.
(Another approach, which doesn't seem to be described in Bott & Tu, is to think of the sets in the cover as the vertices of a simplicial complex, and the $(k+1)$-fold intersections as $k$-simplices; this is called the nerve of the cover. If you then compute the simplicial cohomology of this complex, you recover the de Rham cohomology of the manifold.)
