In the most recent Harvard Qualifying exam, one is asked to prove that if $M$ is a compact oriented manifold, and there are two nonvanishing orientation forms $\xi$ and $\eta$ who's integrals over $M$ agree, then there exists a diffeomorphism $F:M\rightarrow M$ such that $F^*\eta=\xi$
I understand everything but one step in the solution. Let me elaborate, we first show that there exists an $n-1$ form $\sigma$ such that: \begin{align*} d\sigma=\xi-\eta \end{align*} which follows from the fact that the Cohomology class defined by $\xi-\eta$ is zero.
At some point later in the proof, we examine the manifold $\tilde{M}=M\times [0,\infty)$, where the $\mathbb{R}^{\geq 0}$ part of the manifold is represented in coordinates with a $t$. Denote by $\xi$, $\eta$ and $\sigma$ the pull back of $\xi$, $\eta$, and $\sigma$ by the projection $\tilde{M}\rightarrow M$, and define the $n$ form $\omega$ by:
\begin{align*} \omega=(1-t)\xi+t\eta \end{align*} Now the claim is that this one form is closed, but I am not seeing how. In particular, I can write this $n$ form as: \begin{align*} \omega=t\xi -td\sigma \end{align*} Since the pull backs of $\xi$ and $\sigma$ have no $t$ dependence, it follows that:
\begin{align*} d\omega=dt\wedge \xi+td\xi-dt\wedge d\sigma \end{align*} Now $\xi=d\sigma+\eta$ so: \begin{align*} d\omega=dt\wedge \eta+td\xi \end{align*} and $d\xi=d\eta$ so: \begin{align*} d\omega=dt\wedge \eta+td\eta=d(t\eta) \end{align*} which does not look like it is closed to me. Have I made some blunder here?
Edit:
It should indeed by: \begin{align*} \omega=\xi-td\sigma \end{align*} Taking the exterior derivative gives: \begin{align*} d\omega =&d\xi-dt \wedge d\sigma\\ =&d\xi -dt\wedge \xi+dt\wedge \eta \end{align*} but this also does not seem closed.