For example: Let $\mathbb{R}^n$ be a vector space of dimension $n$, then there is a open set $U=\mathbb{R}^2-\{1,2,\ldots,n\}$ such that $$H_{dR}^1(U)\equiv \mathbb{R}^n.$$ Is the following proposition true?
Given a arbitrary $\mathbb{R}$-vector space $V$, is there an open set $U\subset\mathbb{R}^2$ such that $$H_{dR}^1(U)\equiv V.$$
Any hints would be appreciated.
I think that the answer is no (there are vector spaces that are not representing any de Rham cohomology space). I will prove following result: for all manifolds $M$ (as soon as $M$ is second countable) vector spaces $H^n_{dR}(M)$ have cardinality not greater than continuum (maybe there are more natural constraints on theese spaces).
I will use a very simple result: every separable metrizable space is not greater than continuum (see here Cardinality of separable metric spaces for example). Now all we need is to define a metrizable topology on a space $\Omega^n(M)$ of all $n$-forms on $M$.
But elements of $\Omega^n(M)$ are exactly sections of a fiber bundle (exterior $n$-th power of a cotangent bundle) and therefore we can apply a general result (see for example here http://www.staff.science.uu.nl/~crain101/AS-2013/main.pdf p.29). In general the space of smooth sections of a bundle is a nuclear Frechet space (and therefore is always separable).
As for cohomology space - it is a quotient space of a separable Frechet space and therefore it is separable and metrizable.
