I'm studying De Rham Theory on some lecture notes. I have searched in many other references, but I have not found a clear answer. The question is simple, but involves a fundamental definition.
I recall a fact. Let $M$ be an orientable $n$-dimensional manifold without boundary. For all $q\in\mathbb{Z}$, Poincaré Duality provides an isomorphism $H^q_{DR}(M)\simeq(H^{n-q}_c(M))^*$ given by the map $$ \delta: H^q_{DR}(M) \longmapsto (H^{n-q}_c(M))^*,$$ $$[\eta]\longmapsto ([\omega]\mapsto\int_M\eta\wedge\omega). $$
Let $S$ be a $k$-dimensional closed submanifold without boundary embedded in $M$ and let $\int_S\colon H^{k}_c(M)\to\mathbb{R}$ be defined by $\int_S\omega=\int_S \omega_{|S}$. The closed Poincaré dual of $S$ is the unique cohomology class $[\eta_S]\in H^{n-k}_{DR}(M)$ such that, for all $[\omega]\in H^{k}_c(M)$, it holds that $$\int_S\omega=\int_M\omega\wedge \eta_S \tag{1}$$
My question: for which reason we require the closed Poincaré dual of $S$ to satisfy $(1)$ instead of $$\int_S\omega=\int_M\eta_S \wedge \omega,$$ as $\delta$, the Poincaré Duality isomorphism above, would suggest?
It should be true that $\int_M\omega\wedge \eta_S=(-1)^{k(n-k)}\int_M\eta_S \wedge \omega$, so why do we forget the sign?
Is it not important, since it depends on the orientation of $M$?
Is the isomorphism $\delta$ not completely correct?