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?

  • $\begingroup$ How do you obtain $N$? $\endgroup$ – Ryan Budney May 3 at 20:39
  • $\begingroup$ I reformuled my question. Compte on you help. $\endgroup$ – Ady Fall May 3 at 20:43
  • $\begingroup$ Dear Ryan, I hope it's clear how I define the problem. $\endgroup$ – Ady Fall May 3 at 21:27
  • $\begingroup$ 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? $\endgroup$ – Ryan Budney May 3 at 21:44
  • $\begingroup$ 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. $\endgroup$ – 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 $H_k(N)\cong H_k(N,\Cyl \pi)$.

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  • $\begingroup$ 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 $$I : H^{n-k}_{dR}(M, \pi)\longrightarrow H_{k}(N)$$ $\endgroup$ – Ady Fall May 4 at 11:16
  • $\begingroup$ Since this comment is a bit longer I will add it to my answer. $\endgroup$ – Liviu Nicolaescu May 4 at 12:12
  • $\begingroup$ 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 $I(\omega)$? $\endgroup$ – Ady Fall May 4 at 13:24
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    $\begingroup$ This map is not always well defined. The natural map is $$H^{n-k}_{dR}(M)\to H^k(N )\cong\text{Hom}\;\big(\; H_k(N),\mathbb{R}\;\big)$$ $\endgroup$ – Liviu Nicolaescu May 4 at 15:49

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