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There are no exact forms in degree zero. A smooth function has zero exterior derivative precisely when it is locally constant, so $H_{\mathrm{dR}}^0(M)$ is the space of locally constant functions. On a connected nonempty smooth manifold, it is $\mathbb{R}$. An overlapping cover by two connected open subsets therefore does not give the direct-sum formula for de Rham cohomology of a finite disjoint union: restriction sends a constant to equal constants on both pieces.

The two components of a disjoint union are open and closed. Restriction therefore gives an isomorphismIts inverse assigns the first form on $U$ and the second on $V$; it is smooth because there are no overlaps on which compatibility must be imposed. The exterior derivative is local, so this is an isomorphism of cochain complexes.

A form on $M$ is a closed differential form exactly when both restrictions are closed. It is an exact differential form exactly when both restrictions are exact: primitives on $U$ and $V$ join into a global primitive. Taking the quotient therefore yields the de Rham cohomology of a finite disjoint union:The same argument covers degree zero, with the space of degree-minus-one forms taken to be zero.