I've come across two definitions of compactly supported forms in the vertical direction and I'm trying to show they are equivalent. For the setup, let $\pi:E \to M$ be a vector bundle of smooth manifolds and let $\omega$ be a smooth form on $E$.
Definition (1): $\omega$ is a form with compact support in the vertical direction if for every compact subset $K$ of $M$, $\pi^{-1}(K) \cap \text{Supp}\,\omega$ is compact.
Definition (2): $\omega$ is a form with compact support in the vertical direction if $\pi^{-1}(x) \cap \text{Supp}\,\omega$ is compact for every $x \in M$.
The proof for $(1) \implies (2)$ is easy since we take $K = x$. I'm having trouble with the reverse implication. My first tactic was to write $$\pi^{-1}(K) \cap \text{Supp}\,\omega = \bigcup_{x \in K}\pi^{-1}(x) \cap \text{Supp}\,\omega$$ but the right-hand since is not necessarily compact because compactness isn't preserved under arbitrary unions. I then tried using the definition of compactness and picking an open cover for $\pi^{-1}(K) \cap \text{Supp}\,\omega$ and showing it has a finite subcover. The most natural way to do this would seem to be to project the cover into $M$ by $\pi$, find a finite subcover and then take the inverse image but this quickly becomes very messy since $\pi$ is not injective. I'm probably missing something trivial, but I could use some help.
Follow-up: If this is not true, are both definitions used often or is there a "correct" definition to use? I'm reading Bott and Tu and they define compact support in the vertical direction using definition (1), but then only talk about forms satisfying (2) later on. Since (1) implies (2) I'm more asking If we ever need this stronger sense of compactly supported forms in the vertical direction.
They are not equivalent. Here's a counterexample.
Let $E$ be the trivial bundle $\mathbb R\times \mathbb R\to \mathbb R$, with the projection $\pi(x,y) = x$. Let $\omega$ be a smooth real-valued function (i.e., $0$-form) on $E$ that is positive on the set where $x>0$ and $1/x<y<2/x$, and zero elsewhere. (Such a function can be easily constructed using a function that vanishes to infinite order such as $\phi(t) = e^{-1/t}$.)
For each $x\in\mathbb R$, we have $$ \pi^{-1}(x)\cap \operatorname{supp} \omega = \begin{cases} \varnothing, & x\le 0,\\ \{x\} \times [1/x,2/x], & x>0, \end{cases} $$ which is compact in each case. But $\pi^{-1}([0,1])\cap \operatorname{supp} \omega$ is not compact.