push forward of differential form/ integration over fiber

Solution 1:

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.)