My question is on the compact cohomology of $\mathbb{R}$. It is easy to see that $H_c^0(\mathbb{R}) = 0$ because we can't have constant functions on compact support (except for the trivial one). Now we compute $H_c^1(\mathbb{R})$. First, since $\mathbb{R}$ has dimension $1$ then any $1-$form is closed. Now we will search for the exact ones. The idea relies on defining an integral operator $\int_\mathbb{R} : \Omega_c^1(\mathbb{R}) \rightarrow \mathbb{R}$ that simply integrates the form. We claim the following:
The set of exact forms is given by $\ker \int_\mathbb{R}$.
Take $\omega = f(x)dx \in \ker \int_\mathbb{R}$. Let $g(x) = \int_{-\infty}^xf(t)dt$. Then $dg = f(x)dx$ thus an exact form. For the converse, if $\omega = f(x)dx$ is exact then there exists a function $g$ such that $dg = f(x)dx$, i.e. $\frac{\partial g}{\partial x}dx = f(x)dx$. But then $$\int_\mathbb{R}f(x)dx = \int_\mathbb{R}\frac{\partial g}{\partial x}dx = g(b)-g(a) = 0,$$since $g$ has compact support.
Now, the author concludes that $$H_c^1(\mathbb{R}) = \frac{\Omega_c^1(\mathbb{R})}{\ker \int_\mathbb{R}} = \mathbb{R},$$but I don't see why this quotient has dimension one.
Thanks in advance!
Edit with the answer: It relies on the first isomorphism theorem $\int_\mathbb{R} = \frac{\Omega_c^1(\mathbb{R})}{\ker \int_\mathbb{R}} = H_c^1(\mathbb{R})$. Now if one wants to show that $im \int_\mathbb{R}$ has dimension $1$, we must find a compactly supported one form that does not vanishes with integration. We simply take an always positive bump function. Then its integral is positive and it shows that $im \int_\mathbb{R}$ has dimension $1$. So, $$\mathbb{R} \cong im \int_\mathbb{R} \cong = \frac{\Omega^1(\mathbb{R})}{\ker \int_\mathbb{R}} = H^1_c(\mathbb{R}).$$Also, clearly it can't have dimension $2$ since $\Omega^*_c(\mathbb{R}) = 0$ for $* > 1$.
The quotient of $\Omega^1_c(\mathbb{R})$ by the kernel of $\int_\mathbb{R}$ is isomorphic to the image of $\int_\mathbb{R}$. Since the codomain of $\int_\mathbb{R}$ is $\mathbb{R}$, its image can only have dimension $0$ or dimension $1$. To show it has dimension $1$, you just have to find a $1$-form $\omega\in\Omega^1_c(\mathbb{R})$ such that $\int_\mathbb{R}\omega$ is nonzero. You can do this, for instance, with $\omega=f\,dx$ where $f$ is a bump function that is always nonnegative, so its integral must be positive.