Factorial degree obstruction for products of two-spheres ($\mathrm{deg}f=m!{\prod }_{i=1}^m{a}_i$)

Give $(S^2{)}^m$ its product orientation and ${\mathbb{CP}}^m$ its complex orientation. If $u_i$ are the degree-two sphere classes and $h$ the normalized projective class, write $f^{\ast }h={\sum }_i{a}_i{u}_i$. In the source cohomology ring, $u_i^2=0$ and the generators commute. Hence $f^{\ast }{h}^m=m!({\prod }_i{a}_i)u_1\cdots u_m$. Evaluation on the fundamental class gives the displayed mapping degree. Every positive degree is therefore at least $m!$, and the polynomial multiplication map to complex projective space attains that bound.