# An isomorphism in relative De Rham cohomology

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. –  Jonathan Apr 27 '12 at 0:14
Sorry, I actually want the relative cohomology groups with compact support in the vertical direction. I think that fixes it. –  Paul Siegel Apr 27 '12 at 5:17
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 $H^{p-rk(E)}(M)$. –  mland Apr 27 '12 at 9:10

Given a vector bundle $p:E \to M$ of rank $r$, pick some metric on $E$ and use it to define the disc bundle $D \to M$ and unit sphere bundle $S \to M$. An orientation of $E$ gives us a class $\alpha \in H^r(D, S)$ that satisfies the property that for all $x \in M$, the restriction of $\alpha$ to the fiber above $x$ yields a generator of $H^r(D_x, S_x)$. Then we can define a map $H^k(M) \to H^{k+r}(D, S)$ via $\omega \mapsto p^\ast(\omega) \cup \alpha$. The Thom isomorphism theorem tells us that this map is an isomorphism of graded rings. If we use the de Rham model, then elements of $H^\ast(D, S)$ can be represented by differential forms, and the inverse map $H^{k+r}(D,S) \to H^k(M)$ is given by fiberwise integration.
Note that by homotopy invariance, $H^\ast(D, S) \cong H^\ast(E, E^o)$, but I prefer to use $(D,S)$ since it makes the fibers compact.
This is indeed the sort of thing I'm looking for (sorry I got the degrees wrong), but I'm still confused. According to Bott and Tu the Thom class is in the cohomology of $E$ with compact support in the vertical direction rather than in $H^*(D,S)$. Similarly, the target of the Thom isomorphism is $H_{cv}^*(E)$. So implicit in your answer is the assertion that $H^*(D,S) \cong H_{cv}^*(E)$, and it's not clear to me why this is true. Can you elaborate on this point? –  Paul Siegel May 1 '12 at 21:47