Current being cohomologous to a smooth form

Question

Let $\Omega$ be a domain in $\mathbb{R}^{n}$ and $T$ be a current of degree p on $\Omega$. Define the $p$-th cohomology group for currents as $H^{p}_{cur}(\Omega,\mathbb{R}):=\frac{\{real-valued~closed~p-currents~on~\Omega\}}{\{real-valued~exact~p-currents~on~\Omega\}}.$ Since $\mathcal{C}^{\infty}~p$-forms are also $p$-currents, we say a current $T$ is cohomologous to a form $\alpha$ if $[T]_{cur}=[\alpha]_{cur}$. Now for a compact orientable manifold $X$ without boundary, the map $\pi:H^{p}_{DR}(X,\mathbb{R})\mapsto H^{p}_{cur}(\Omega,\mathbb{R})$ given by $\pi([\alpha])=[\alpha]_{cur}$ becomes an isomorphism.(Note that $H^{p}_{DR}(X,\mathbb{R})$ denotes the p-th De-Rham cohomology group and $[\alpha]$ denotes De-Rham cohomology class of the closed form $\alpha$). In particular, every closed p-current is cohomologous to a closed smooth p-form(follows from the surjectivity of $\pi$). Is this still valid for $\Omega?$