How do I compute the DeRham Cohomology of $H^{1}(\Bbb{R})$ and $H^{1}(S^{1})$ ?
Like for $H^{1}(\Bbb{R})$,
$H^{1}(\Bbb{R}) = \frac{Z^{1}(\Bbb{R})}{B^{1}(\Bbb{R})}$, But what will be $Z^{1}(\Bbb{R})$ ? , where $Z^{1}(\Bbb{R}) = \{$ 1 - forms which are closed like $M dx = 0 \}$.
and
$B^{1}(\Bbb{R})= \{$ 1 -forms which are exact $\}$.
After this I am unable to proceed?
(1) By the fundamental theorem of calculus $B^1(\mathbb{R})=Z^1(\mathbb{R})=\Omega^1(\mathbb{R})$, hence $H^1(\mathbb{R})=0.$
(2) Show that $\int\colon Z^1(S^1)=\Omega^1(S^1)\rightarrow\mathbb{R}$ has $B^1(S^1)$ as kernel, then use the first isomorphism theorem (in linear algebra) to compute $H^1(S^1)\cong\mathbb{R}$.