# Poincare duality

\section{Список литературы}

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 $$$$z \sim z^{\prime} \Longleftrightarrow z = z^{\prime} \quad \text {or} \quad (z, z^ {\prime} \in \partial M \quad \text{and} \quad \pi(z) = \pi (z^{\prime})).$$$$ We define the topological space $$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$$ can be not smooth sometimes. How to define the map $$I : H^{n-k}_{dR}(M)\longrightarrow H_{k}(N)$$ when $$F$$ is not a singleton?

• As it stands, the answer is ... You don't. May 3, 2020 at 18:56
• Try using the de Rham theorem: $H^*_{dR}(M) \cong H_{sing}^*(M;\mathbb{R})$ May 3, 2020 at 18:58
• So there's no map between them , even in terms of integral? May 3, 2020 at 18:58