= Solution
For finite-dimensional <Complex projective space> $\mathbb{CP}^r$, use the filtration by its linear projective subspaces. The difference $\mathbb{CP}^j\setminus\mathbb{CP}^{j-1}$ is $\mathbb C^j$, so there is one <topological cell> in every even dimension $0,2,\ldots,2r$, and none in odd dimensions. Every cellular boundary vanishes for degree reasons. The integral <cohomology> is therefore $\mathbb Z$ in those even degrees and zero elsewhere.
For the <cohomology ring>, let $x\in H^2(\mathbb{CP}^r;\mathbb Z)$ be the <Poincare dual> of a <projective hyperplane>, using the complex <orientation>. Generic hyperplanes meet transversely, and a collection of $j$ of them intersects in $\mathbb{CP}^{r-j}$. The geometric interpretation of the <cup product> identifies $x^j$ with the dual of that intersection. Restricting to a linear $\mathbb{CP}^j$, the $j$ hyperplanes meet in one point, with positive sign because the transverse normal directions are complex lines. Hence
$$
\left\langle x^j,[\mathbb{CP}^j]\right\rangle=1.
$$
Thus $x^j$ is a primitive generator in degree $2j$, not merely a nonzero multiple. There are no classes beyond degree $2r$, so
$$
\boxed{H^*(\mathbb{CP}^r;\mathbb Z)\cong\mathbb Z[x]/(x^{r+1}),\qquad |x|=2.}
$$
The point case $r=0$ has ring $\mathbb Z$. Taking the union of the finite projective spaces gives $H^*(\mathbb{CP}^{\infty};\mathbb Z)=\mathbb Z[x]$, since each fixed cohomological degree stabilizes. This proves the <cohomology ring of complex projective space> computation.
For the <rank bound for a nonsingular complex bilinear map>, write $a=\dim_{\mathbb C}U$, $b=\dim_{\mathbb C}V$, $R=\operatorname{im}\phi$, and $s=\dim_{\mathbb C}R$. Assume $a,b\geq1$. Injectivity with either nonzero factor fixed implies $\phi(u\otimes v)\ne0$ whenever $u,v\ne0$. Therefore
$$
F:\mathbb{CP}^{a-1}\times\mathbb{CP}^{b-1}\longrightarrow\mathbb{CP}^{s-1},\qquad ([u],[v])\longmapsto[\phi(u\otimes v)]
$$
is well defined and continuous. Scaling either vector only scales the output, and in local projective charts the formula is given by bilinear coordinates with no simultaneous zero. Let $h$ be the target's degree-two generator and $x,y$ the pullbacks of the two factor generators. The <Künneth theorem> gives
$$
H^*(\mathbb{CP}^{a-1}\times\mathbb{CP}^{b-1};\mathbb Z)=\mathbb Z[x,y]/(x^a,y^b).
$$
Fixing $[v]$ or $[u]$ makes $F$ a projective linear embedding induced by an injective <linear map>. Such an embedding pulls back the hyperplane class to the factor's hyperplane class. Since the two restrictions determine $H^2$ of the product, $F^*h=x+y$, with the generator of a point factor interpreted as zero.
Put $d=a+b-2$. In the source <cohomology ring>, every term in $(x+y)^d$ vanishes except the one with exponents $a-1,b-1$, and
$$
F^*(h^d)=(x+y)^d=\binom{a+b-2}{a-1}x^{a-1}y^{b-1}\ne0.
$$
The binomial coefficient is a positive integer and the final monomial generates the top integral <cohomology>. If $s\leq d$, the class $h^d$ in $\mathbb{CP}^{s-1}$ would be zero. This contradiction proves
$$
\boxed{\operatorname{rank}\phi\geq\dim_{\mathbb C}U+\dim_{\mathbb C}V-1.}
$$
When $a=b=1$, the nonzero pure-tensor image already gives $s\geq1$. A positivity convention is genuinely necessary: if $U=0$, $V=\mathbb C^2$, and $W=0$, all restrictions to the zero tensor subspaces are injective, but the literal bound would read $0\geq1$. The displayed theorem is therefore the nonzero-factor version. If a nonzero factor is infinite-dimensional, either factor restriction already bounds the image dimension below by that factor dimension; the sum of two positive cardinal dimensions, with at least one infinite, is their maximum, giving the corresponding cardinal bound.
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