Suppose $M$ is a n-dimensional, finite type, oriented, smooth manifold. A $k$-dimensional cycle in $M$ is a pair ($S$,$\phi$), where $S$ is a compact, oriented $k$-dimensional manifold without boundary, and $\phi$: $S$$M$ is a smooth map. A $k$-cycle ($S$, $\phi$) defines a linear map $H^k(M)\rightarrow R$ given by $w \mapsto \int_S \phi^* w$ where $w \in H^k(M)$ and $\phi^*$ is the pullback. In other words, that each cycle defines an element in $(H^k(M))^*$. Via the Poincaré duality we identify the vector space $(H^k(M))^∗$ with $H^{n−k}_{cpt}(M)$ where $H^{n-k}_{cpt}(M)$ denotes the compactly supported cohomology group. Thus there exists $\delta_S \in H^{n−k}_{cpt}(M)$, such that \begin{align} \int_M w \wedge \delta_S=\int_S \phi^*w. \end{align} The compactly supported cohomology class $\delta_S$ is called the Poincare dual of the pair $(S, \phi)$.

My question is that: since in the above formula uses the wedge product, it is more or less related to De Rham cohomology(not?). Can the above formula be rephrased into the language of singular cohomology? More explicitly, consider a triangulation $\Gamma$ of $M$, and a $k$-cycle ($S,\phi)$ can be represented by a singular $k$-cycle, denoted as $C_S$ and the cohomology class $w$ can also defined in the singular cohomology sense, so we still use $w$ to denote it. Then I try to rephrase the above formula as \begin{align} C_S \cap w= \Gamma \cap (w\cup w_S) \end{align} or equivalently \begin{align} \int_{C_S} w= \int_{\Gamma} w\cup w_S \end{align} where $w_S \in H^{n-k}(\Gamma)$ is the Poincaré dual of $C_S$ in $\Gamma$, so that $w \cup w_S \in H^n(\Gamma)$, parallel to $\delta_S$ defined above. $\cap$ is the cap product and $\cup$ means the cup product. Do this rephrase make sense? Is this $w_S$ right to be Poincaré dual of $C_S$?


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