Compactly supported cohomology of a connected manifolds of finite type.

Question

Is it possible that the compactly supported cohomology (over rational number filed) of connected manifold (non-oriented) of finite type is zero? In case of ordinary cohomology at least Zero-th cohomology group is non-zero. In case of oriented manifold this zero-th cohomology groups corresponds to the n-th cohomology group of compactly supported cohomology (using Poincare duality). Let us supposed M be a non-compact non-oriented manifold. It is true or not there is at least one cohomology group of compactly supported cohomology of M is non-zero. If yes then what is its degree?

Answer 1

No, for instance if $M$ is the open Mobius band, we have $H^*_c(M;\Bbb Q) = 0$.

The point is that if $M$ is the interior of a compact manifold with boundary $\bar M$, then $H^*_c(M) \cong H^*(\bar M, \partial \bar M)$; and then there is the isomorphism $$H^*(\bar M, \partial \bar M) \cong H^*(\bar M/\partial \bar M, *),$$ where we take on the right side relative cohomology at the basepoint we collapsed $\partial \bar M$ into. Cohomology of a locally contractible space relative to a basepoint is just reduced cohomology.

Now the closed mobius band $\bar M$ has boundary a circle, and collapsing it to a point is homeomorphic to filling it in with a disc, which is the real projective plane $\Bbb{RP}^2$. So $H^*_c(M;\Bbb Q) = \tilde H^*(\Bbb{RP}^2;\Bbb Q) = 0$.