# Thom isomorphism for cohomology with compact support

Let $p \colon E \to M$ be a smooth real vector bundle of rank $n$, $R$-oriented for a commutative ring $R$.

Recall that the Thom isomorphism theorem states that there is a unique class $u \in H^n(E, E \setminus M)$ such that for all $x \in M$, the restriction $$H^n(E, E \setminus M;R) \to H^n(E_x, E_x \setminus \{0\};R)$$ maps $u$ to the prefered generator given by the orientation. And furthermore $$H^k(M;R) \to H^{k+n}(E, E \setminus M;R), \qquad \alpha \mapsto p^*(\alpha) \cup u$$ is an isomorphism.

Question: According to exercise 6.20 in Differential Forms in Algebraic Topology by Bott & Tu (p.65), there is also an isomorphism $$H_c^k(M;R) \xrightarrow{\cong} H_c^{k+n}(E;R)$$ in cohomology with compact support given by the same formula. Can this be concluded from the above usual Thom theorem?

• I think the two cases (with compact support or not) are separated. The proofs should be similar, but you need to "do it again" in all cases. If I recall correctly, even Poincare lemma's proof for $H^{*}_{c}$ is different from $H^{*}$ in Bott&Tu. The difference can be subtle. – Bombyx mori Sep 8 '16 at 14:17
• Does this even work with general coefficients besides $\mathbb{R}$ when the approach by Bott & Tu is to work with differential forms? I'm thinking about the de Rham theorem. – hallborrey Sep 8 '16 at 15:03
• I meant that you may need to prove the statement all-over again as an execrise. As far as I know, there does not exist a Thom isomorphism theorem for compact supported functions when the base field is $\mathbb{C}$ (say, using Dolbeault cohomology or something) in literature. But maybe this can be proved. The case over $\mathbb{Z}$ is proved in Milnor, and it is not difficult. – Bombyx mori Sep 8 '16 at 15:28