To what extent is the global angular form well-defined?

I am reading Differential Forms in Algebraic Topology by Bott & Tu. The authors constructed the global angular form $$\psi$$ for an oriented $$k$$-sphere bundle $$E$$ over a smooth manifold $$M$$. It has the properties:

1. $$\psi$$, when restricted to each fiber, is the generator for the top cohomology of the fiber. To be more precise, $$\psi|_{E_p}$$ is a closed $$k$$- form and $$\int_{E_p}\psi|_{Ep}=1$$. The integral makes sense since the sphere bundle is oriented.
2. $$d\psi=-\pi^*(e)$$, where $$\pi:E\to M$$ is the natural projection and $$e=e(E)$$ is the Euler class of the sphere bundle.

The global angular form is clearly not uniquely characterized by the above two properties. For any closed form $$\eta$$ on $$M$$, $$\psi+\pi^*(\eta)$$ will satisfy 1 and 2 if $$\psi$$ does. Note that $$\psi$$ is in general not closed, hence does not represent a cohomology class. Now my question is:

To what extent is the global angular form of an oriented sphere bundle unique, satisfying 1 and 2 above?

Since the global angular form is in some sense choosing a polar coordinate system for each fiber and the Euler class indicates the obstruction against this. (I am sorry for my imprecision at this point. I may need some help here.) So I have conjectured the following proposition:

With notations as above. Let $$\omega$$ be a $$k$$-form on $$E$$ such that

1. $$\omega$$ restricts to the generator of the top cohomology of each fibre and
2. $$d\omega=\pi^*(\eta)$$ for some closed form $$\eta$$ on $$M$$.

Then $$\eta$$ is the Euler class of the sphere bundle.

Can someone prove of disprove this proposition? Any comment on the intuitive understanding of the global angular forms is appreciated. Thanks!