Independence of $H(f)=\int_M \alpha \wedge f^* \beta$ on choice of $d\alpha=f^*\beta$? I came across the following UCLA qual question while studying for my upcoming qual:

Let $f: M^{4n-1} \to N^{2n}$ be a smooth map between closed connected oriented manifolds of the indicated dimensions and $\beta$ any $2n$-form on $N$ such that $\int_N \beta=1$. Suppose $f^*\beta$ is exact: $f^* \beta = d \alpha$.
(a) Show that $H(f) = \int_M \alpha \wedge f^* \beta$ is independent of the choices of $\alpha$ and $\beta$.
(b) Show that if $f$ and $g$ are homotopic, $H(f)=H(g)$.

(http://www.math.ucla.edu/~yingkun/UCLAtopoqual.pdf, bottom of page 11)
My computations are showing that  $\alpha \wedge f^*\beta$ is exact, so Stokes' Theorem would imply that $H(f)$ is always zero on a boundaryless manifold such as $M$. This makes part (b) silly, too.  Either way, I haven't been able to make any sense of the hypothesis that $\int_N \beta=1$. Any thoughts?
 A: HINTS: 
First argue it doesn't depend on the $\alpha$ with $d\alpha = f^*\beta$. You should end up with needing to know that the wedge product of a closed form and an exact form is exact.
Second, because of the condition that $\int_N \beta = 1$, any other such form $\beta'$ will differ from $\beta$ by an exact form, say $d\eta$. In that case you can give an explicit choice of $\alpha'$ with $f^*\beta' = \alpha'$. Along the way, it should become relevant that when $k>2n$, the only $k$-form on $N$ is $0$.
A: Update, Dec '15: I'm moving the full argument (completed using Ted's hints) from the original post to here for aesthetic/legibility reasons.
Proof for Part (a). 
Given $\alpha' \in \Omega^{2n-1}(M)$ also satisfying $d\alpha'=f^*\beta$, we first observe that
\begin{align*}
d(\alpha' \wedge \alpha) &= d\alpha' \wedge \alpha +(-1)^{2n-1} \alpha' \wedge d\alpha \\
&= (-1)^{(2n)(2n-1)} \alpha \wedge d\alpha' - \alpha' \wedge d\alpha \\
&= \alpha \wedge f^*\beta - \alpha' \wedge f^*\beta.
\end{align*} 
An application of Stokes' Theorem gives
\begin{equation*}
\int_M \alpha \wedge f^*\beta -\int_M \alpha' \wedge f^*\beta  = \int_M d(\alpha' \wedge \alpha) = \int_{\partial M=\emptyset} \alpha' \wedge \alpha = 0.
\end{equation*}
Now suppose $\beta' \in \Omega^{2n}(N)$ also satisfies $\int_N \beta'=1$. Both $\beta$ and $\beta'$ are closed because they are top-dimensional forms. Furthermore, they represent nontrivial cohomology classes since exact forms integrate to zero on the boundaryless manifold $N$. Thus $[\beta]$ and $[\beta']$ both generate $H^{2n}(N;\mathbb{R})\cong \mathbb{R}$. Since $\int_N \beta = \int_N \beta'$, we in fact have $[\beta]=[\beta']$, i.e. $\beta-\beta'=d\theta$ for some $(2n-1)$-form $\theta$. This gives
\begin{align*}
\int_M \alpha \wedge f^* \beta - \int_M \alpha \wedge f^*\beta' &= \int_M \alpha \wedge f^* d\theta \\
&= \int_M \big(d\alpha \wedge f^* \theta - d(\alpha \wedge f^*\theta)\big)\\
&= \int_M d\alpha \wedge f^*\theta - \int_{\partial M=\emptyset} \alpha \wedge f^*\theta\\
&= \int_M f^* \beta \wedge f^*\theta -0\\
&= \int_M f^*(\beta \wedge \theta)\\
&=0,
\end{align*}
where the last line follows from the fact that the $(4n-1)$-form $\beta \wedge \theta$ on the $2n$-manifold $N$ is necessarily zero. QED
Proof for Part (b).
Let $F: M \times [0,1] \to N$ be a homotopy with $F_0=f$ and $F_1=g$.  It suffices to show that 
$\qquad$ (i) $F^*\beta$ is exact, i.e. $F^*\beta=d\theta$ for some $\theta \in \Omega^{2n-1}(M\times I)$, and
$\qquad$ (ii)  $\theta \wedge F^*\beta$ is closed,
as this will imply
\begin{equation*}
0 = \int_{M \times [0,1]} d(\theta \wedge F^*\beta) = \int_{M} \theta_1 \wedge F_1^*\beta - \int_M \theta_0 \wedge F^*_0 \beta = H(g)-H(f).
\end{equation*}
To prove (i), first note that $F$ itself is (smoothly) homotopic to the map  $\tilde f: M \times [0,1] \to N$ defined by $\tilde f(x,t)=f(x)$, hence they induce the same homomorphism $H^{2n}(M \times I;\mathbb{R}) \to H^{2n}(N;\mathbb{R})$.  Since $[\tilde f^* \beta]$ maps to  $[f^* \beta]=0$  under the isomorphism  $H^{2n}(M \times I;\mathbb{R}) \to H^{2n}(M;\mathbb{R})$, it follows that $[F^*\beta]=[\tilde f^*\beta]=0$. This implies that $F^*\beta$ is exact, i.e. we can write $F^*\beta= d\theta$ for some $\theta \in \Omega^{2n-1}(M\times I)$.
We demonstrate (ii) directly:
\begin{align*}
d(\theta \wedge F^*\beta)&=d\theta \wedge F^*\beta + (-1)^{2n-1} \theta \wedge d(F^*\beta) \\
&= F^*\beta \wedge F^*\beta - \theta \wedge F^*(d\beta)
\\&=F^*(\beta \wedge \beta)\\
&=0.
\end{align*}
Thus $H(g)=H(f)$. QED
