\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?
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.