De Rham cohomology disjoint union of cylinders

Question

Could someone explain me (sorry for the, maybe, trivial question :-) ) how to prove that $H^1_{DR}((S^1\times\mathbb{R}) \sqcup(S^1\times\mathbb{R}))=\mathbb{R}^2$?

I'm talking about the de Rham cohomology.

Answer 1

You need only the following three facts:

• $H^1(X\sqcup Y) = H^1(X) \oplus H^1(Y)$,

• $H^1 (X\times \mathbb R) = H^1(X)$, and

• $H^1(\mathbb S^1) = \mathbb R$.

The first and the third facts can be proved directly, while the second one (sometimes referred as the Poincare lemma) might be a bit harder to show. Everything are well discussed in the book Differential forms in Algebraic topology by Bott and Tu.