Let $M$ be be a principal $S^1$-bundle, i.e. a smooth manifold with a smooth $S^1$-action (multiplication to the right) that has no fixed points such that the orbits through every point $p \in M$, that is, $$\mathcal{O}_p = \{p \cdot z \ \mid z \in S^1 \} $$ are the fibers of a submersion $h:M \to B$, where $B$ is another smooth manifold.
Let us define the vector field $V$ on $M$ as such: $$\forall p \in M: V_p = \frac{d}{dt}\bigg\rvert_{t = 0} p \cdot e^{2\pi it}. $$ (I have also defined it in one of my previous questions - Flow of vector field $V_p = \frac{d}{dt} \left|_{t=0} \right. \ p \cdot e^{2\pi it}$ on a manifold $M$ with an $S^1$-action).
Now let us call a differential form $\omega$ on $M$ horizontal if $i_V(\omega) = 0$, equivariant if $R_\lambda^*\omega = \omega$ for every $\lambda \in S^1$, where $R_\lambda(p):= p \cdot \lambda$, and basic if it is both. Trivial properties of these forms are the following: if $\omega$ is equivariant, then $dw$ is also equivariant; $\omega$ is equivariant if and only if $\mathcal{L}_V(\omega) = 0$, and if $\omega$ is basic then $dw$ is also basic.
I want to prove two things: firstly: $(i)$ that $h^*$ is a (linear) isomorphism between the space of differential forms on $B$ and the space of all basic forms on $M$, and secondly: $(ii)$ if we take a differential $1$-form on $M$ such that $\omega(V) = 1$ and take $\eta$ such that $dw = h^*\eta$ (which exists by $(i)$), then $\eta$ is a closed form and the de Rham cohomology class of $\eta$ (let us denote it by $[\eta]$) is independent of the choice of $\omega$.
For $(i)$, I am quite certain that injectivity of $h^*$ follows from the properties of the pullback, since if $\omega$ and $\eta$ are differential $k$-forms on $B$ such that $$h^*\omega = h^*\eta, $$ then for every vector fields $X^1, \cdots, X^k \in \mathfrak{X}(M)$, we have that $\forall p \in M$, \begin{equation} \omega_{h(p)}((dh)_p(X^1_p), \cdots, (dh)_p(X^k_p)) = \eta_{h(p)}((dh)_p(X^1_p), \cdots, (dh)_p(X^k_p)), \ \ \ (1) \end{equation} so if we want to prove that for some vector fields $Y^1, \cdots Y^k \in \mathfrak{X}(B)$, $$\omega(Y^1, \cdots, Y^k) = \eta(Y^1, \cdots, Y^k), $$ we can choose the surjectivity of $dh$ to choose some vector fields such that $(1)$ is satisfied.
However, I am having trouble proving that $h^*$ is surjective and that it actually is well-defined, by which I mean that $h^*\eta$ is a basic form on $M$ when $\eta$ is a differential form on $B$, as I don't know how to begin.
For $(ii)$, we see that $\eta$ is closed because $h^*$ is an isomorphism from $(i)$ and because the pullback commutes with the de Rham operator. However, I am also not sure how to prove that the cohomology class of $\eta$ is independent of $\omega$.
$h^*$ is well defined since for every element $g\in S^1$, $h\circ g=h$. This implies that for every $p$-form $\alpha$ defined on $B$, $g^*(h^*\alpha=(h\circ g)^*\alpha=h^*\alpha$.
For the second question, remark that $i_V\omega=1$ implies that $L_V\omega=0$. If $\omega_1(V)=\omega_2(V)=1$, we deduce that $L_V(\omega_1-\omega_2)=0$ and $\omega_1-\omega_2$ is basic, thus there exists $\alpha$ such that $\omega_1-\omega_2=h^*\alpha$. We deduce that if $h^*\eta_i=d\omega_i, i=1,2$. $\eta_1-\eta_2=d\alpha$ since the restriction of $h^*$ to basic forms is exact.