Existence of differential form which satisfies Integration along the fibres

Question

Hi I have a question here, but it has quickly become an unfocused mess. I also believe I have answered the main question I had in edit $1$, but I now I have a new problem about trying to tackle the problem a different way. Assume all coordinate covers are compatibly oriented.

Let $\pi:E\rightarrow M$ be a compact oriented smooth fibre bundle, with compact oriented fibre $F$, and compact oriented base (essentially just making everything as nice as possible). And define integration along the fibres to be the map: \begin{align*} \pi_*:\Omega^{n}(E)\longrightarrow \Omega^{n-f}(M) \end{align*} (here $f$ is the dimension of the fibre) by assigning an $n$ form $\omega$ to the $n-f$ form on $M$ which satisfies: \begin{align*} \int_M\beta\wedge \pi_*\omega=\int_{E}\pi^*\beta\wedge \omega \end{align*} for all $\beta\in \Omega^{m+f-n}(M)$. Showing this map is well defined, i.e. if $\eta$ satisfies the property above, then $\eta=\pi_*\omega$, however showing that $\pi_*\omega$ exists in the first place is a more difficult task.

Here is what I attempt, suppose that $(U_i,\psi_i)$ is a cover of $M$ which trivializes $E$, and further that each $U_i$ is a coordinate chart, so there exist diffeomorphisms $\phi_i$ for each $U_i$ such that $\phi(U_i)$ is open in $\mathbb{R}^m$. It follows that we can obtain a coordinate cover for $E$, by taking a coordinate cover $(Y_j,\theta_j)$ and constructing coordinate charts by: \begin{align*} \phi_{ij}=(\phi_i\times \theta_j)\circ \psi_i:\psi_i^{-1}(U_i\times Y_j)\longrightarrow \phi_i(U_i)\times \theta_j(Y_j) \end{align*} and going forward, we denote $\psi_i^{-1}(U_i\times Y_j)$ by $V_{ij}$. Locally we can then use coordinates $(x^1,\dots, x^m, y^1,\dots, y^m)$, and write any $n$ form in coordinates as a linear combination of forms of the type: \begin{align*} \phi_{ij}^{-1*}(\alpha)=(\pi\circ \phi_{ij}^{-1})^*(\xi)\wedge f(x,y) dy^{i_1}\wedge \cdots \wedge dy^{i_k} \end{align*} where $\xi\in \Omega^{k-n}(M)$ , and of the type: \begin{align*} \phi_{ij}^{-1*}(\eta)=(\pi\circ \phi_{ij}^{-1})^*(\xi)\wedge f(x,y)dy^1\wedge\cdots \wedge dy^f \end{align*} So we define $\pi_*$ on forms of the type as identically zero, and on forms of the second type locally by: \begin{align*} \pi_*(\phi_{ij}^{-1*}\eta)=\xi\int_{\theta_j(Y_j)}f(x,y)dy^1\wedge \cdots \wedge dy^n \end{align*} So then choose a partition of unity $\rho_{ij}$ subordinate to the cover $(V_{ij},\phi_{ij})$, and define $\pi_*$ globally by: \begin{align*} \pi_*(\omega)=\sum_{ij}\phi_i^*(\pi_*(\phi_{ij}^{-1*}(\rho_{ij}\omega))) \end{align*}

Essentially what I think this does it gives us a bunch of forms defined on $U_i\times V_j$, writes them as forms on some open subset of $\mathbb{R}^m\times \mathbb{R}^f$, integrates out the $\mathbb{R}^f$, then pulls the form on $\mathbb{R}^M$ back to $U_i$. This seems to be the clearest way of defining this procedure, but working with it is rather unwieldy, in particular, I am struggling to show that this satisfies the property described earlier. I think I am maybe making this too complicated...

Anyways, let $\beta\in\Omega^{m+f-n}(M)$, then: \begin{align*} \int_E\pi^*(\beta)\wedge \omega=&\sum_{ij}\int_{\phi_{ij}(V_{ij})}\phi_{ij}^{-1*}(\pi^*(\beta)\wedge \rho_{ij}\omega)\\ =&\sum_{ij}\int_{\phi_i(U_i)\times \theta_j(Y_j)}(\pi\circ \phi_{ij}^{-1})^*(\beta)\wedge \phi_{ij}^{-1}(\rho_{ij}(\omega)) \end{align*} Fix $i$ and $j$ then: \begin{align*} \phi_{ij}^{-1}(\rho_{ij}(\omega))=\sum_k\sum_{i_1<\cdots< i_k}(\rho_{ij}\circ \phi_{ij}^{-1})(\pi\circ \phi_{ij}^{-1})^*(\xi_k)\wedge f_{i_1\cdots i_k}dy^{i_1}\wedge \cdots \wedge dy^{i_k} \end{align*} where $\xi_k\in\Omega^{n-k}(M)$. However: \begin{align*} \beta\wedge \xi_k\in\Omega^{m+f-k}(M) \end{align*} so unless $k=f$ the whole expression in the integrand is $0$. It follows that the only non zero part is then: \begin{align*} \phi_{ij}^{-1}(\rho_{ij}(\omega))=(\rho_{ij}\circ \phi_{ij}^{-1})(\pi\circ \phi_{ij}^{-1})^*(\xi)\wedge f(x,y)dy^{1}\wedge \cdots \wedge dy^{f} \end{align*} for some $\xi\in \Omega^{n-f}$, hence: \begin{align*} \int_{\phi_i(U_i)\times \theta_j(Y_j)}(\pi\circ \phi_{ij}^{-1})^*(\beta)\wedge \phi_{ij}^{-1}(\rho_{ij}(\omega))=&\int_{\phi_i(U_i)\times \theta_j(Y_j)}(\pi\circ \phi_{ij}^{-1})^*(\beta\wedge \xi)\wedge (\rho\circ \phi_{ij}^{-1})f(x,y)dy^1\wedge \cdots \wedge dy^n \end{align*} Now note that: \begin{align*} \pi\circ \phi_{ij}^{-1}=&(\pi\circ \psi_i^{-1})\circ (\phi_i\times \theta_j)^{-1}\\ =&\pi_{U_i}\circ (\phi_i\times \theta_j)^{-1}\\ =&\phi_{i}^{-1}\circ \pi_{\phi(U_i)} \end{align*} where $\pi_{U_i}:U_i\times Y_j\rightarrow U_i$, and $\pi_{\phi(U_i)}:\phi(U_i)\times \theta_j(Y_j)\rightarrow \phi(U_i)$ are the projections. We can thus rewrite the integral as: \begin{align*} \int_{\phi_i(U_i)\times \theta_j(Y_j)}(\pi\circ \phi_{ij}^{-1})^*(\beta)\wedge \phi_{ij}^{-1}(\rho_{ij}(\omega))=&\int_{\phi_i(U_i)\times \theta_j(Y_j)}(\phi_{i}^{-1}\circ \pi_{\phi_i(U_i)})^*(\beta\wedge \xi)\wedge (\rho\circ \phi_{ij}^{-1})f(x,y)dy^1\wedge \cdots \wedge dy^n \end{align*} And I think I can then rewrite this as: \begin{align*} \int_{\phi_i(U_i)\times \theta_j(Y_j)}(\pi\circ \phi_{ij}^{-1})^*(\beta)\wedge \phi_{ij}^{-1}(\rho_{ij}(\omega))=&\int_{\phi_i(U_i)}(\phi_{i}^{-1})^*(\beta\wedge \xi)\int_{\theta_j(Y_j)} (\rho\circ \phi_{ij}^{-1})f(x,y)dy^1\wedge \cdots \wedge dy^n\\ =&\int_{U_i}(\beta\wedge \xi)\phi_i^*\left(\int_{\theta_j(Y_j)} (\rho\circ \phi_{ij}^{-1})f(x,y)dy^1\wedge \cdots \wedge dy^n\right)\\ =&\int_{U_i}\beta\wedge \phi_i^*(\pi_*(\phi_{ij}^{-1*}(\rho_{ij}\omega))) \end{align*} Since this holds for all $i$ and $j$, it follows that: \begin{align*} \sum_{ij}\int_{\phi_i(U_i)\times \theta_j(Y_j)}(\pi\circ \phi_{ij}^{-1})^*(\beta)\wedge \phi_{ij}^{-1}(\rho_{ij}(\omega))=&\sum_{ij}\int_{U_i}\beta\wedge \phi^*((\pi_*(\phi_{ij}^{-1*}(\rho_{ij}\omega)))\\ =&\int_M\beta\wedge \pi_*(\omega) \end{align*} so the defining $\pi_*$ as above works, implying such a form exists.

What I am worried about is mainly separating the integrals, and getting rid of the $\pi^*$, but I feel like that should be fine since we're integrating over $\mathbb{R}^m\times \mathbb{R}^f$. I am also worried about the final step, since my partition of unity is buried inside the definition of $\pi_*(\omega)$, and I'm not really sure if that's ok...

Any advice or help on showing that this form exists (perhaps a simpler argument?) would be greatly appreciated.