 Solution

The area form $\sigma$ on $S^2$ is closed. For any closed two-form $B$ on a base, the twisted cotangent symplectic form $\omega +{\pi }^{\ast }B$ is closed and nondegenerate: pairing a putative kernel vector first with vertical vectors kills its horizontal component, and then pairing with horizontal vectors kills its vertical component. Thus both ${\omega }_{+}$ and ${\omega }_{-}$ are symplectic.

The cotangent bundle deformation-retracts onto its zero section. Their cohomology classes satisfyThey are therefore not strongly isotopic.

Let $f:S^2\to S^2$ be an orientation-reversing isometry. Then $f^{\ast }\sigma =-\sigma$, while its cotangent lift preserves the canonical form and satisfies $\pi \circ f_{\#}=f\circ \pi$. Consequentlyso the two forms are symplectomorphic.