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\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 \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. How to define the map $I : H^{n-k}_{dR}(M)\longrightarrow H_{k}(N)$ when $F$ is not a singleton?

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    $\begingroup$ As it stands, the answer is ... You don't. $\endgroup$ May 3, 2020 at 18:56
  • $\begingroup$ Try using the de Rham theorem: $H^*_{dR}(M) \cong H_{sing}^*(M;\mathbb{R})$ $\endgroup$
    – William
    May 3, 2020 at 18:58
  • $\begingroup$ So there's no map between them , even in terms of integral? $\endgroup$
    – Ady Fall
    May 3, 2020 at 18:58
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    $\begingroup$ 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 $ X $ - the base $F$- the fiber are smooth compact manifolds without boundary. By sending $\partial M$ to $X$ we obtained a new manifold called $N$. This manifold can be not smooth. How to define a map $I : H^{n-k}_{dR}(M)\longrightarrow H_{k}(N) when N is not smooth $\endgroup$
    – Ady Fall
    May 3, 2020 at 19:36
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    $\begingroup$ OK, now the definitions are clear but the question is not. For instance, what if I tell you to use zero map, would you object? You probably want some naturality properties. However, if $F$ is a singleton, $N=M$, so you are then asking for a natural map $H^{n-k}(M)\to H_k(M)$ (where, probably, $n=\dim(M)$). The Poincare duality gives you $H^{n-k}(M)\to H_k(M, \partial M)$. There are no "natural" maps from that to $H_k(M)$: The natural map goes in the opposite direction. $\endgroup$ May 3, 2020 at 20:49

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I can come up with at least a partial answer, using the de Rham theorem (dR) and Poincare-Lefschetz duality (PLD). Note that $X$ is naturally embedded as a subspace of $N$, as the image of $\partial M$ under the quotient map $q\colon M \to N$. Then we have $$H^{n-k}_{dR}(M) \stackrel{dR}{\cong} H^{n-k}(M;\mathbb{R}) \stackrel{PLD}{\cong} H_k(M,\partial M;\mathbb{R}) \stackrel{q_*}{\to} H_k(N, X; \mathbb{R}).$$

If $H_{k+1}(X) = H_{k-1}(X) = 0$ then $H_k(N,X)\cong H_k(N)$ so we can naturally make a homomorphism which maps to $H_k(N)$, but I'm not sure how to do so in general unless $H_*(X;\mathbb{R})$ vanishes in the right degrees.

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  • $\begingroup$ I have tried using the fundamental class of $N$. Since $M$ and $N$ have the same dimension, but am not sure if it's a good definition. Let $[N]=\sum_{j}c_{j} $the fundamental class of $N$. We define the following \begin{eqnarray*} \mu=\left[N\right]\frown: H^{n-k}_{dR}(M,\pi) & \longrightarrow & H_{k}(N) \\ \omega & \longmapsto & \sum\limits_{j}\left(\int_{c_{j}\mid_{\Delta^{n-k}}}\omega\right)c_{j}\mid_{\Delta^{k}} \end{eqnarray*} $\endgroup$
    – Ady Fall
    May 3, 2020 at 22:45
  • $\begingroup$ William, how to define the map if $ H_{*}(X,\mathbb{R})$ vanishes in the right degrees. Could you give an example of definition in that case, please $\endgroup$
    – Ady Fall
    May 3, 2020 at 22:51

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