# Intuitive interpration of the Thom isomorphism and relative cohomology

Let $p: E \rightarrow B$ be an an oriented real vector bundle or rank $n$. Then there exists a unique class $u \in H^n(E, E - B; \mathbb{Z})$, where $B$ is embedded into $E$ as the zero section, such that for any fiber $F$ the restriction of $u$ to $F$ is the class induced by the orientation of $F$. $H^k(E) \cong H^{k+n}(E,E - B)$ and this isomorphism is given explicitly by $\, x \mapsto x \cup u$. I would like to get a feel for this isomorphism. For the record: I am only really interested in smooth vector bundles over manifolds.

If $k+n$ is greater than the dimension of the base space then I think this is easily interpreted (though please correct me if I say something wrong or misleading). Writing out the the long exact sequence associated with the relative cohomology, we see that $H^{k+n}(E)$ is zero as $E$ deformation retracts to $B$. We see that in this case $H^{k+n}(E;E - B) \cong H^{k+n-1}(E - B)$. The isomorphism $H^k(E) \cong H^{k+n}(E,E - B)$ then tells us about the cohomology of $H^{k+n-1}(E - B)$, which I can "feel" is the twisting of the vector bundle.

For the case $k+n$ less than the dimension of the manifold, it is less clear for me how to think of relative cohomology.

I know that Bott and Tu have some interpretation like this and a lot of people swear by that book, but to be honest that book just confused me. Maybe I should try again... Any other ideas?

2. By excision, that relative cohomology is the same as the disk bundle relative its boundary. In that case, we can think about it, with coefficients in $\mathbb{R}$, as having to do with forms on the disk bundle that vanish on the boundary. We are looking for a form which integrates to $1$ on each fiber. There is always a choice of such a thing when the bundle is trivial, but when we glue, we might worry about a $1$ becoming a $-1$. An orientation says we can make choices that do not cancel out. Once we have such a thing, cupping with it is an isomorphism with inverse given by "integrating it out."