 Cup power

For a cohomology class $a$, its rth cup power is its r-fold product in the cohomology ring. Naturality gives $f^{\ast }(a^r)=(f^{\ast }a{)}^r$. A nilpotent class has some positive cup power zero; the nilpotence order is preserved by ring isomorphisms and can obstruct a homotopy equivalence or a nonzero mapping degree.