In Bott & Tu's "Differential Forms in Algebraic Topology" on page 50 & 51, the authors define the Poincare dual of a submanifold as follows:
Let $M$ be an oriented smooth manifold of dimension $n \in \mathbb{N}$ and $i: S \rightarrow M$ a smooth submanifold of dimension $0 \leq k \leq n$ which is closed. Here the authors specify that "closed" means closed as a topological subspace of $M$. Once can now show that pulling back a compactly supported form on $M$ along the inclusion gives a compactly supported form on $S$. Let $\omega \in \Omega^k_c (M)$ be closed, then since $S$ is orientable the assignment $\omega \mapsto \int_S i^*(\omega)$ gives a well defined linear functional $\int_S i^*(-): \Omega^k_c (M) \rightarrow \mathbb{R}$.
Now the authors state that "by Stokes's Theorem, integration over $S$ induces a linear functional on $H^k_c(M)$". But I fail to see why this is.
We need to check that the functional vanishes on exact $k$-forms. So let $d \omega \in \Omega^k_c (M)$ be exact. Then
$$ \int_S i^*(d \omega) = \int_S d (i^*(\omega)) = \int_{\partial S} i^*(\omega)$$
In the first equality we apply that pullbacks commute with the differential while in the second we apply Stokes's theorem. However, I fail to see why the latter integral is $0$. If the boundary of $S$ was empty, this would be the case, but we only assume $S$ to be closed as a topological subspace, not as a manifold (i.e. compact and empty boundary).
For example, this question assumes $S$ to have empty boundary. Could it be that Bott & Tu assume $S$ to be closed (in the stronger sense of compact and no boundary). That would also imply that $S$ is closed as a topological subspace.
Thank you very much in advance.