How does the induced map on homology work on compact manifolds with boundary? Suppose $u \colon \Sigma^p \to M^n$ is a smooth map between compact smooth manifolds.  For the moment, suppose that both $\Sigma$ and $M$ have no boundary.  Let $[\Sigma] \in H_p(\Sigma)$ denote the fundamental class of $\Sigma$.  In all that follows, we'll use $\mathbb{R}$ coefficients for homology and cohomology.
Let $\omega \in \Omega^p(M)$ be a closed $p$-form on $M$, and let $[\omega] \in H^p(M)$ denote its (de Rham) cohomology class.  Then
\begin{equation*}
\int_\Sigma u^*\omega = \langle [u^*\omega], [\Sigma] \rangle = \langle [\omega], u_*[\Sigma] \rangle \tag{$\star$}
\end{equation*}
where $\langle \cdot, \cdot \rangle$ is the pairing between homology and cohomology, and $u_* \colon H_p(\Sigma) \to H_p(M)$.

Question: Suppose now that $\Sigma$ has non-empty boundary, but $M$ is still without boundary.  I guess we should interpret $[\Sigma] \in H_p(\Sigma, \partial \Sigma)$.  My questions are:
(1) How does the induced map $u_*$ on homology work, and more importantly:
(2) Does equation $(\star)$ still hold true?

I'm aware that this is a very basic question, but references to textbooks that address this would be appreciated nonetheless.
 A: There is an induced map in relative homology for a map of pairs $(X,A) \to (Y,B)$. for any set $A \subset M$ with $u(\partial \Sigma) \subset A$, you get an induced map $$H_*(\Sigma, \partial \Sigma) \to H_*(M, A).$$ 
The most common geometric scenario of interest like this is $A = \partial M$ and $u$ a boundary-preserving map, which you have excluded in your hypotheses. You could of course just use $A = u(\partial \Sigma)$, but computing this relative homology seems difficult and is of course wildly dependent on both $u$ and $\Sigma$.
The thing that relative homology $H_*(M,A)$ pairs with is relative cohomology $H^*(M,A)$; the easiest description to think of is Godbillon's as forms which vanish on $A$ (I think this requires $A$ to be a point-set closed submanifold for this to be a well-behaved theory). Then your pairing formula works exactly as expected, the point being that Shifrin's $\int_{u_*\partial \Sigma} \eta$ is tautologically zero, as $\eta \big|_{A} = 0$ and $u(\partial \Sigma) \subset A$.
