Jim's comment is right, the glueing information is hidden in the map $\mathbb R^2 \to \mathbb R^2$ corresponding to the map $\phi : H^2_c(U\cap V) \to H^2_c(U) \oplus H^2_c(V)$.
Recall that if $U$ and $V$ are opens of $\mathbb R^2$,
if we have a map $\iota : \mathbb U \to \mathbb V$, it induces a map $\iota^* : \Omega^2_c(\mathbb U) \to \Omega^2_c(\mathbb V)$,
such that $\iota^* (f dxdy) = g dxdy$ where $g$ satisfies $f = (g \circ \iota) * J$ where $J$ is the jacobian of $\iota$, and $g=0$ outside the image of $\iota$.
Consequently, depending on the (non-changing) sign of $J$, we have $\int_U \omega = \pm \int_V (\iota^*(\omega))$ forall $\omega \in \Omega^2_c(U)$ (the change of variable formula is exactly what we have, but with an absolute value on the jacobian).
Therefore you have to keep track wether all your inclusion maps preserve orientation or not.
Write $U \cap V = W = W_1 \cup W_2$, so that $H^2_c(U\cap V) = H^2_c(W_1) \oplus H^2_c(W_2)$.
We know that $U,V,W_1,W_2$ are diffeomorphic to $\mathbb R^2$, so the isomorphisms are induced by $\alpha : \omega \in \Omega^2_c(U) \mapsto \int_U \omega$
$\phi$ is given by $\phi(\omega_1 \oplus \omega_2) = (\iota_{W_1 \to U}^*(\omega_1) - \iota_{W_2 \to U}^*(\omega_2), \iota_{W_1 \to V}^*(\omega_1) - \iota_{W_2 \to V}^*(\omega_2))$.
So $\alpha \circ \phi (\omega_1 \oplus \omega_2) = (\int_U \iota_{W_1 \to U}^*(\omega_1) - \int_U \iota_{W_2 \to U}^*(\omega_2), \int_V \iota_{W_1 \to V}^*(\omega_1) - \int_V \iota_{W_2 \to V}^*(\omega_2))$.
In the Möbius case, we usually pick maps such that $\iota_{W_1 \to U},\iota_{W_2 \to U},\iota_{W_1 \to V}$ are orientation preserving, and $\iota_{W_2 \to V}$ is orientation-reversing, so we obtain :
$\alpha \circ \phi (\omega_1 \oplus \omega_2) =
(\int_{W_1} \omega_1 + \int_{W_2} \omega_2, \int_{W_1} \omega_1 - \int_{W_2} \omega_2)$
thus the corresponding map $\tilde{\phi} : \mathbb R^2 \to \mathbb R^2$ is $(x,y) \mapsto (x+y,x-y)$, which is an isomorphism. Therefore, its kernel and cokernel, $H^1_c(M)$ and $H^2_c(M)$, are both zero.
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.
Best Answer
Using Poincaré duality $$H_c^k \simeq H_{2-k}$$ you can reduce the computation to this answer.