# Poincare duality-differential geometry

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$$ can be not smooth sometimes. The pair $$(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?

• How do you obtain $N$? – Ryan Budney May 3 at 20:39
• I reformuled my question. Compte on you help. – Ady Fall May 3 at 20:43
• Dear Ryan, I hope it's clear how I define the problem. – Ady Fall May 3 at 21:27
• Is the locally-trivial bundle $M$ or its boundary? And you have not said what the map $I$ is supposed to be. Is there a situation where your map has a name? – Ryan Budney May 3 at 21:44
• the boundary is the locally trivial bundle. I can want to define any in case where $F$ is not a singleton. Assume that is $F$ is a singleton, then projection pi is trivial and in that case $M=N$. I suppose now the case where $F$ is not a singleton. Then $N$ may be not smooth. I wanna define a map (any map) which holds when is not smooth. – Ady Fall May 3 at 21:47

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 $$H_k(N)\cong H_k(N,\Cyl \pi)$$.
• Thanks Liviu. But how can I define exiplicitly the map $I$ when N is not smooth? In the case where $N$ is not smooth, how can I define a map $$I : H^{n-k}_{dR}(M, \pi)\longrightarrow H_{k}(N)$$ – Ady Fall May 4 at 11:16
• Thanks Liviu. thanks very much. So, could you define the map $I$ please. hint $\omega \in H^{n-k}_{dR}(M)$. How to sent $\omega$ to $H_{k}(N)$. what is equal $I(\omega)$? – Ady Fall May 4 at 13:24
• This map is not always well defined. The natural map is $$H^{n-k}_{dR}(M)\to H^k(N )\cong\text{Hom}\;\big(\; H_k(N),\mathbb{R}\;\big)$$ – Liviu Nicolaescu May 4 at 15:49