Poincare duality-differential geometry

Question

Let $ M $ be a smooth and compact manifold with boundary $\partial M = X \times F $ on which the structure of a smooth locally trivial bundle $$ \pi: \partial M \longrightarrow X $$ where the $ X $ and the fiber $ F $ are smooth compact manifolds without boundary. Consider the equivalence relation on the set M \begin{equation} z \sim z^{\prime} \Longleftrightarrow z = z^{\prime} \quad \text {or} \quad (z, z^ {\prime} \in \partial M \quad \text{and} \quad \pi(z) = \pi (z^{\prime})). \end{equation} We define the topological space $ N = M / \sim $ as the quotient space of the manifold M with respect to the equivalence relation above. Informally speaking, $ N $ is obtained from $ M $ (by contracting the fibers of the bundle $ \pi $ to points). The set $ N $ is a disjoint union $ N = X \sqcup M^{\circ} $ of the manifold $ X $ and the interior $ M^{\circ} $ of $ M $. The natural projection of $$ p: M \longrightarrow N $$ coincides with the identity map on $ M ^ {\circ} $ and the projection $ \pi $ on $ \partial M $. So the manifold $N$ can be not smooth sometimes. The pair $(M,\pi)$ is called a manifold with fibered boundared. How to define the map $I : H^{n-k}_{dR}(M,\pi)\longrightarrow H_{k}(N)$ when $F$ is not a singleton?

Answer 1

Note that $N$ is homeomorphic to the union of $M$ with $\DeclareMathOperator{\Cyl}{Cyl}$ the mapping cylinder $\Cyl(\pi)$ of the bundle projection $\newcommand{\pa}{\partial}$ $\pi:\pa M\to X$. Denote by $M^\circ$ the interior of $M$.

Observe that the Poincare Duality for $M^\circ$ (or equivalently for $(M,\pa M)$) implies $$ H^{n-k}_{dR}(M)\cong H^k_{dR}(M^\circ) \cong H^k_{cpt}(M^\circ). $$ The extension by $0$ defines a morphism
$$ H^k_{cpt}(M^\circ)\to H^k_{cpt}(N)\cong H^k(N)\cong \mathrm{Hom}\big(H_k(N),\mathbb{R}\big). $$ Comment. Above I assumed that $H^\bullet_{dR}(M,\pi)=H^\bullet_{dR}(M)$. Now observe that $$ H^{n-k}_{dR}(M)\cong H_k(M,\pa M). $$ From the excision property of homology we deduce that the inclusion $$ (M,\pa M)\hookrightarrow (N,\Cyl \pi) $$ induces an isomorphism $$ H_k(M,\pa M)\cong H_k(N,\Cyl \pi). $$ If $H_k(X)=H_{k-1}(X)=0$, then $H_k(N)\cong H_k(N,\Cyl \pi)$.