TITLE: Calculating the cohomology with compact support of the open Möbius strip
QUESTION [26 upvotes]: I am having problems calculating the cohomology with compact support of the open Möbius strip (without the bounding edge).
I am using the Mayer Vietoris sequence: U and V are two open subsets diffeomorphic to $\mathbb{R}^2$ and $U\cap V$ is diffeomorphic to two copies of $\mathbb{R}^2$.
$H^0_C(M)=0$ and that's ok, but then I get the exact sequence 
$ 0 \rightarrow H^1_C(M) \rightarrow H^2_C(U\cap V) \rightarrow H^2_C(U)\oplus H^2_C(V) \rightarrow H^2_C(M)\rightarrow 0$
where both $H^2_C(U\cap V)$ and $H^2_C(U)\oplus H^2_C(V)$ have dimension 2. 
I would say that the function $\delta:H^2_C(U\cap V) \rightarrow H^2_C(U)\oplus H^2_C(V)$ in the exact sequence sends $(\phi,\psi)$ to $(-j_U(\phi + \psi),j_V(\phi + \psi))$ where $\phi$ and $\psi$ are generators of $H^2_C(U\cap V)$.
So I would say that $Im\ \delta$ is one dimensional and spanned by $(-j_U(\phi + \psi),j_V(\phi + \psi))$ and $\dim H^1_C(M)=\dim H^2_C(M)=1$. But, I have read that all compact cohomology classes of the Möbius strip are zero. So I must be wrong somewhere.

REPLY [7 votes]: This is one of the execrises from Bott-Tu. Here I try to write a more friendly version of mercio's answer. Note that the Mayer-Vietoris sequence of forms with compact support is induced by the short exact sequence
$$
0\leftarrow \Omega^{*}_{c}(M)\leftarrow \Omega_{c}^{*}(U)\oplus \Omega_{c}^{*}(V)\leftarrow \Omega_{c}^{*}(U\cap V)\leftarrow 0
$$
where the first map is $(u,v)\rightarrow u+v$, second map $(\omega)\rightarrow (-j_{*}\omega, j_{*}\omega)$, where $j: \Omega_{c}^*(U)\rightarrow \Omega_{c}^{*}(M)$ is the inclusion map. 
Now back to the setting of Mobius band, we have two pieces $U,V$ each differeomorphic to $\mathbb{R}^{2}$. The induced map is given by
$$
H^{0}_{c}(U\cap V)\rightarrow H^{0}_{c}(U)\oplus H^{0}_{c}(V)\rightarrow H^{0}_{c}(M)\rightarrow H^{1}_{c}(U\cap V)\rightarrow H^{1}_{c}(U)\oplus H^{1}_{c}(V)\rightarrow H^{1}_{c}(M)\rightarrow H^{2}_{c}(U\cap V)\rightarrow H^{2}_{c}(U)\oplus H^{2}_{c}(V)\rightarrow H^{2}_{c}(M)
$$
By Poincare lemma we know that $H^{0}_{c}(U\cap V)=H^{0}_{c}(U)=H^{0}_{c}(V)=0$. Thus $H^{0}_{c}(M)=0$. Now after truncation we have
$$
0\rightarrow H^{1}_{c}(M)\rightarrow H^{2}_{c}(U\cap V)\rightarrow H^{2}_{c}(U)\oplus H^{2}_{c}(V)\rightarrow H^{2}_{c}(M)\rightarrow 0
$$
where the middle map is given by $\omega=(\omega_1, \omega_2)\rightarrow (-(j_{U}(\omega), (j_{V})(\omega))$. We use the identification of the Mobius band to let 
$$
j_{U}(\omega_1)=\omega_1, j_{U}(\omega_2)=\omega_2, j_{V}(\omega_{1})=\omega_{1}, j_{V}(\omega_{2})=-\omega_{2}
$$
Then the image is given by $(-\omega_1-\omega_2, \omega_1-\omega_2)$ is isomorphic to $\mathbb{R}\oplus \mathbb{R}$. As a result both the kernel and cokernel are zero. Thus $H^{*}_{c}(M)=0$.