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Let $E$ be a smooth oriented vector bundle over a smooth manifold $M$ and let $E^0$ be the complement of the zero section in $E$. I would like a reasonably explicit isomorphism between the relative De Rham cohomology group $H^p(E,E^0)$ and $H^p(M)$. Perhaps this is some version of the Thom isomorphism?

In case anyone needs a refresher, $H^*(E,E^0)$ is the cohomology of the complex whose chain groups are $\Omega^p(E,E^0) := \Omega^p(E) \oplus \Omega^{p-1}(E^0)$ and whose differential is given by $d(\omega_1, \omega_2) = (d \omega_1, i^* \omega_1 - d \omega_2)$.

EDIT: I actually want to prove that $H_{cv}^P(E,E^0) \cong H^p(M)$, where $H_{cv}^p(E,E_0)$ is the cohomology of the complex described above where all forms have compact support in the vertical direction.

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    I'm a bit confused. If $E \to M$ is a vector bundle, then $M$ is a retract of $E$ and they have the same cohomology. It seems to me that the relative cohomology $H^\ast(E, E^o)$ should be isomorphic to the (reduced?) cohomology of the Thom space of $E$, which is certainly not isomorphic to $H^\ast(M)$ in general.2012-04-27
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    Sorry, I actually want the relative cohomology groups with compact support in the vertical direction. I think that fixes it.2012-04-27
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    are you sure you do not expect an index shift in your cohomology? I suppose that your $H_{cv}^p(E,E^0)$ is isomorphic to singular cohomology with compact support (at least if $M$ is compact) $H^p_{c}(E,E^0) \cong \tilde{H}^p(M(E))$ but by the Thom isomorphism this is the shifted cohomology $H^{p-rk(E)}(M)$.2012-04-27

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