push forward of differential form/ integration over fiber It is elementary that differential forms can be pulled back via a smooth map between manifolds. However, I was reading a paper and came across a construction about push forward of a differential form via a submersion which I didn't fully understand.
The paper pointed to Differential Forms in Algebraic Topology by Bott and Tu for reference. However, since I have little background in algebraic topology, I would like to know if anyone can show me a more detailed explanation, or point me to some references.
Below is the construction as described in the paper:

If $f: X\rightarrow Y$ is a submersion from an oriented manifold of dimension $n$ to an oriented manifold of dimension $m \leq n$. Then the fibers are manifolds of dimension $r=n-m$.

So far this is OK, and it continues:

Integration over the fibers gives a map $f_*: D^p(X)\rightarrow D^{p-r}(Y)$ defined as follows.
Any $p$-form $\phi$ on $X$ with compact support can be written $\phi = \psi \wedge f^*\omega$, where $\psi$ is an $r$-form with compact support on $X$ and $\omega$ is a $(p-r)$-form on $Y$. To see this, use a partition of unity to write $\phi$ as a sum of forms with support in a coordinate neighborhood, and in local coordinates the decomposition becomes obvious.
We can then consider $f_*\psi$ on $Y$ with compact support defined by $f_*\psi (y)=\int_{f^{-1}(y)} \psi$ and define $f_*\phi = f_* \psi \wedge \omega$.

What I didn't understand is how the $p$-form $\phi$ on $X$ can be decomposed as $\psi \wedge f^*\omega$. (Even though it says it's obvious in local coordinates...)
Is this decomposition unique? If not, then the push forward $f_* \phi$ better not depend on the decomposition..?
 A: The decomposition is not unique, because we can replace $\psi \wedge f^{\ast} \omega$ with $(f^{\ast}(g) \psi) \wedge f^{\ast} (g^{-1} \omega)$ where $g$ is some smooth function $Y \to \mathbb{R}_{>0}$. Of course, this doesn't effect the pushforward, since $\int_{f^{-1}(y)} f^{\ast}(g) \psi = g(y) \int_{f^{-1}(y)} f^{\ast}(g)$ and then $\left( \int_{f^{-1}(y)} f^{\ast}(g) \psi ) (g^{-1} \omega) \right)= g(y) g(y)^{-1} \omega = \omega$. More generally, if $\phi = \alpha \wedge f^{\ast}(\beta \wedge \gamma \wedge \delta)$ with $\alpha$ an $r-k$ form, $\beta$ and $\gamma$ $k$-forms and $\delta$ a $p-r-k$ form, then $\phi = (\alpha \wedge f^{\ast} \beta) \wedge f^{\ast}(\gamma \wedge \delta) = (-1)^k (\alpha \wedge f^{\ast} \gamma) \wedge f^{\ast}(\beta \wedge \delta)$ are two such factorizations.
I think the simplest way to prove indepedence of the decomposition is to describe the pushforward in terms of its integrals. For $\phi$ any compactly supported $p$-form on $X$, and $B$ any embedding of a closed $(p-r)$-dimensional ball into $Y$, we have $\int_B f_{\ast} \phi = \int_{f^{-1}(B)} \phi$ (using Fubini). Since a $(p-r)$-form is defined by its integrals over all $(p-r)$-dimensional balls, this proves uniqueness of such a $f_{\ast} \phi$ (but not existence, use Bott and Tu's argument for that.)
