Suppose $\pi: Y \rightarrow X$ is a smooth submersion of oriented manifolds, $\dim X = n$, $\dim Y = m$. Then there exists a unique operator of integration along fibers, denoted $\pi_*$, taking compactly supported $l$-forms on $Y$ ($l \geq m-n$) to compactly supported $l$-forms on $X$, given by:
$$ \int_Y (\pi^* \beta) \wedge \mu = \int_X \beta \wedge (\pi_* \mu) $$
for any $\beta \in \Omega^{m-l}(X)_0$ and $\mu \in \Omega^l(Y)_0$.
I am trying to see why if $\phi:Y \rightarrow Y$ and $\psi:X \rightarrow X$ are proper orientation preserving maps satisfying $\pi \circ \phi = \psi \circ \pi$, then:
$$ \pi_* \circ \phi^* = \psi^* \circ \pi_* $$
I am stuck beyond applying the definition:
$$ \int_X \beta \wedge \pi_* (\phi^* \mu)
= \int_Y \pi^* \beta \wedge (\phi^* \mu)
$$
How to treat $\psi^* \circ \pi_*$ and use uniqueness of $\pi_*$ to reach the equality?
Edit: If $\phi$ and $\psi$ are orientation-preserving diffeomorphisms I have:
$$
\int_X \psi^* \beta \pi_* \phi^* \mu =
\int_Y \pi^* \psi^* \beta (\phi^* \mu) =
\int_Y \phi^* \pi^* \beta (\phi^* \mu) =
\int_Y \phi^* ( \pi^* \beta \mu) =
\int_Y \pi^* \beta \mu =
\int_X \beta \pi_* \mu
$$
Then maybe I can use somehow that $\psi^*$ is an isomorphism?
Best Answer
I still need to correct your misstatement at the beginning: If $\mu$ is an $l$-form on $Y$, then $\pi_*\mu$ is an $\big(l-(m-n)\big)$-form on $X$; the rest is correct.
We do need to use the fact that $\phi$ and $\psi$ are orientation-preserving diffeomorphisms. Then we'll have $\int_X\psi^*\alpha = \int_X \alpha$ for any $n$-form $\alpha$ on $X$ and $\int_Y\phi^*\beta = \int_Y\beta$ for any $m$-form $\beta$ on $Y$. (You used the latter in your argument at the end, and need one more step with the former.) Since I find it hard to read your argument without wedges, I will rewrite it: For an arbitrary form $\beta$ on $X$, we have \begin{align*} \int_X \psi^*\beta\wedge \pi_*\phi^*\mu &= \int_Y \pi^*\psi^*\beta\wedge\phi^*\mu = \int_Y \phi^*(\pi^*\beta\wedge\mu) \\ &\overset{\dagger}= \int_Y \pi^*\beta\wedge\mu = \int_X \beta\wedge\pi_*\mu \overset{\dagger}=\int_X\psi^*(\beta\wedge\pi_*\mu) \\ &=\int_X\psi^*\beta\wedge \psi^*\pi_*\mu. \end{align*} We've used the diffeomorphism property at the two marked equalities. Since $\beta$ (and therefore $\psi^*\beta$, once again since $\psi$ is a diffeomorphism) is arbitrary, the desired equality follows.
Despite what you said, we do need to have diffeomorphisms (or close to it). Here's a simple example. Take $\pi\colon S^1\to \{\text{pt}\}$, and let $\phi(z)=z^2$. (You can replace the trivial space with any manifold you want and just take the product with $S^1$.) Let $\mu=d\theta/2\pi$. Then $\pi_*\phi^*\mu = \pi_*(2\mu) = 2 \ne 1 = \pi_*\mu$.