to prove such a statememnt without an explicit classification result and computations. However, we note the following.

\[thm:ray\] Let $Z$ and $Z'$ be two symplectic deformation equivalent varieties with second Betti number $2$. Assume that there is a cohomology class $A \in H^2(Z, {{\mathbb}{Q}})=H^2(Z', {{\mathbb}{Q}})$ which is an ample class for both $Z$ and $Z'$. Then any extremal ray of $Z$ (resp. $Z'$) lie in the effective cone of $Z'$ (resp. $Z$) and at least one of the extremal rays stays extremal.

There are several situations where the assumptions are satisfied, e.g. when $(Z, \omega)$ and $(Z', \omega')$ are symplectomorphic, or symplectic isotopy to each other (and the combination of the two). The idea of the proof comes from a result of Wi[ś]{}niewski [@DefNefValue]. In fact, all one need to do is to adapt the proof to our case.

It is easy to see that if all the extremal rays stay effective under symplectic deformations, then any variety $W$ symplectic deformation equivalent to such a $Z$ (assuming $b_2(Z)=2$) has the same extremal ray under the identification $H_2(Z, {{\mathbb}{Q}}) \cong H_2(W, {{\mathbb}{Q}})$. Hence the minimal model program (MMP) will produce the same fibration structure (at least in dimension $4$, c.f. Lemma 