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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.

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    I think you are probably computing for an open annulus. Once you take orientation data into account, you ought to get something like $\phi\mapsto \phi+\psi$ and $\psi\mapsto \phi-\psi$, which will be two dimensional.2012-01-26
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    it makes sense to me, but you mean $\delta(\phi)=(-\phi,\phi)$, $\delta(\psi)=(-\psi,-\psi)$?2012-01-26
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    Yes that's right.2012-01-27
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    Can anyone answer this question decisively? I do not find the answer below persuasive.2012-06-18

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