de Rham cohomology with compact support isomorphic to $\mathbb{R}$ The $k$-dimensional de Rham cohomology with compact support is defined as $$H_c^k(M)=\frac{Z_c^k(M)}{B_c^k(M)} $$
where $Z_c^k(M)$ is the vector space of all closed $k$-forms with compact support on $M$ and $B_c^k(M)$ is the vector space of all $k$-forms $d\eta$ where $\eta$ is a $(k-1)$-form with compact support on $M$. A Comprehensive Introduction To Differential Geometry states that for a connected orientable manifold $M$ (I think $M$ is considered as with $\partial M=\emptyset $) we have $$H_c^n(M)\simeq \mathbb{R} \quad \quad \quad (*)$$
Then he says that this assertion is equivalent to two other assertion, which I list below: 


*

*Given a fixed $\omega$ such that $\int_M\omega\neq0$, then for any $\omega'$ n-form with compact support there exists a real number $\lambda$ such that $\omega' - \lambda \omega $ is exact (It is not mentioned, but I assume both $\omega$ and $\omega'$ are supposed to be closed)

*An closed form $\omega$ is the differential of another form with compact support if $\int_M \omega=0$
I wanted to prove this (the equivalence of this assertions), but I cannot see how this is true. 
Any help is welcome 
Thanks in advance 
 A: First, there is an error in your statement (1): it should say $\omega'-\lambda\omega$ is exact, not $\omega-\lambda\omega'$.  (Also, $n$ should presumably be the same as the $k$ you mention elsewhere, the dimension of $M$.)
Now there is a linear map $I:H^k_c(M)\to\mathbb{R}$ which maps $[\omega]$ to $\int_M\omega$ (note that this is well-defined because the integral of an exact form is $0$).  Since there exists a compactly supported $k$-form with nonzero integral, $I$ is surjective.  Since $\mathbb{R}$ is a 1-dimensional vector space, the following are then equivalent:


*

*(a) $I$ is an isomorphism.

*(b) $H^k_c(M)$ is 1-dimensional, i.e. $H^k_c(M)\cong \mathbb{R}$

*(c) $\ker I$ is trivial.


But (c) is exactly the same as your statement (2), since an element of $\ker I$ is $[\omega]$ such that $\int\omega=0$ and to say $[\omega]=0$ means that $\omega$ is the differential of a form with compact support.  And of course (b) is the same as your statement (*).
It remains to be shown that your statement (1) is also equivalent to the others.  Note first that (1) says exactly that $[\omega]$ spans $H^k_c(M)$, since it says that for any $[\omega']\in H^k_c(M)$, there is some scalar $\lambda$ such that $[\omega']=[\omega]$.  So, (1) implies $H^k_c(M)$ is 1-dimensional.  Conversely, assume statement (2).  Fix $\omega$ such that $\int_M \omega\neq 0$ and take any other $\omega'$, and let $\lambda=\frac{\int_M\omega'}{\int_M\omega}$.  Then $\int_M \omega'-\lambda\omega=0$, so $\omega'-\lambda\omega$ is exact.
