# De Rham's theorem for top-forms in manifolds with boundary

In page 79 of Bott-Tu, "Differential Forms in Algebraic Topology", they define the relative de Rham theory as follows:

Let $$f:S\to M$$ be a smooth map. Define the complex $$\Omega^*(f)$$ by $$\Omega^k(f):=\Omega^k(M)\oplus\Omega^{k-1}(S)$$ $$\underline{\mathrm{d}}(\alpha,\beta)=(\mathrm{d}\alpha,f^*\alpha-\mathrm{d}\beta)$$ It is easy to prove that $$\underline{\mathrm{d}}^2=0$$ which allows us to define the cohomology $$H^*(f)$$. As a particular case, one can consider a submanifold $$\imath:N\hookrightarrow M$$ and define $$\Omega^*(M,N):=\Omega^*(\imath)$$

My interest lies in the case when $$N=\partial M$$ and $$M$$ compact, where one can also define the integral of top forms as $$\int_{(M,\partial M)}(\alpha,\beta):=\int_M\alpha-\int_{\partial M}\beta$$ It is easy to check, using Stoke's theorem, that $$\int_{(M,\partial M)}\underline{\mathrm{d}}(\alpha,\beta)=0$$ Thus, we have a well defined map $$\tag{1}\label{one}\int_{(M,\partial M)}:H^n(M,\partial M)\to\mathbb{R}$$

If $$\partial M=\varnothing$$, then $$H^n(M,\partial M)=H^n(M)$$ and the previous integral is the standard one. The de Rham's theorem for top-forms then tells us that if $$M$$ has no boundary $$\tag{2}\label{two}\int_M:H^n(M)\to\mathbb{R}\quad \text{ is an isomorphism}$$ However, with boundary we have:

1. It is surjective (applying \eqref{two} over the boundary and using elements of the form $$(0,\beta)$$).

2. Its kernel is isomorphic to $$H^n(M)$$. Sketch of the proof: for every $$[\alpha]\in H^n(M)$$, build an element $$[(\alpha,\beta)]$$ such that $$\int_{(M,\partial M)}(\alpha,\beta)=0$$ using de Rham's theorem over the boundary. This map is well defined.

I have a heuristic argument to show that $$H^n(M)$$ is always zero: given $$\alpha\in\Omega^n(M)$$, take the double of $$M$$ along the boundary $$\partial M$$ and extend to some $$\widetilde{\alpha}\in\Omega^n(M\sqcup_{\partial M}M)$$ such that its integral is zero (using a tubular neighborhood over $$\partial M$$). Then using \eqref{two} (the double has no boundary) shows that $$\widetilde{\alpha}$$ is exact and, therefore, its pullback to $$M$$, which is $$\alpha$$, is also exact.

This seems a very strong result that I haven't found anywhere, while the proof seems very simple, thus I doubt if there are obstructions to the extension that invalidate the proof.

So the questions I have in mind (all of them are almost the same question) are:

1. Is $$H^n(M)=0$$ if $$M$$ is compact with boundary?
2. Is there a useful characterization of $$H^n(M,\partial M)$$ that can be used in this context?
3. Is there a de Rham's theorem like \eqref{two} for manifolds with boundary (with no prescribed boundary conditions)?
4. Is there a de Rham's theorem like \eqref{two} for relative cohomology?
5. If $$H^n(M)\neq 0$$, is there another a map $$G:H^n(M,\partial M)\to \mathbb{R}$$ such that $$(\int_{(M,\partial M)},G)\to\mathbb{R}^2$$ is an isomorphism?

It is indeed true that $$H^n(M)=0$$ if $$M$$ is a compact manifold with boundary. In particular, $$H^n(M,\partial M)\cong\mathbb{R}$$ by Lefschetz duality (as Chris Gerig mentioned) and the integral (1) is an isomorphism.