Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 114 / 3 / c / Solution

For nonzero finite-dimensional $U$ and $V$, the intended assertion is true. Write $a=\dim U$, $b=\dim V$, and replace $W$ by $\mathrm{im}\varphi$, of dimension $r$. Injectivity on the stated slices implies $\varphi (u\otimes v)\ne 0$ whenever $u,v\ne 0$. Consequently the bilinear map defines a continuous map of Complex projective spacesThe rank $r$ is at least one, since one such nonzero tensor has nonzero image.

Let $L_U,L_V,L_W$ be the respective tautological bundles. Fiberwise, $\varphi$ identifies $L_U\otimes L_V$ with $F^{\ast }{L}_W$. For the positive hyperplane classes $x=c_1(L_U^{\ast })$, $y=c_1(L_V^{\ast })$ and $z=c_1(L_W^{\ast })$ (zero when the projective space is a point), the first Chern class of a tensor product of complex line bundles givesThe Künneth theorem, together with part (a), identifies the product cohomology ring withThere are no Tor terms because the factor groups are free. In particular, its monomials $x^i{y}^j$ with $0\le i<a$, $0\le j<b$ form an integral additive basis. In top degree,If $r\le a+b-2$, the target relation $z^r=0$ would imply $(x+y{)}^r=0$ and hence contradict this nonzero top power. ThusWhen $a=b=1$, the already-established inequality $r\ge 1$ gives the same conclusion. The complex bilinear dimension bound is sharp: multiplication of complex polynomials of degrees less than $a$ and $b$ has target dimension $a+b-1$, is injective in either nonzero fixed factor, and its image spans every monomial in that target.

Literal zero-space qualification. The printed assertion does not explicitly exclude zero vector spaces. If $U=0$, $V={\mathbb{C}}^2$ and $W=0$, its slice-injectivity hypothesis is vacuous, while the claimed inequality would be $0\ge 1$. Thus, with zero spaces permitted, this is a counterexample to the assertion exactly as written; the proof above supplies the usual nonzero finite-dimensional interpretation.