Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 114 / 3

The standard CW complex structure on ${\mathbb{CP}}^n$ has one cell in every even dimension from $0$ to $2n$, and ${\mathbb{CP}}^k$ is its $2k$-skeleton. Collapsing that subcomplex leaves one zero-cell and one cell in each dimension $2i$ for $k+1\le i\le n$. All cellular boundaries vanish, so

Let $x\in H^2({\mathbb{CP}}^n;\mathbb{Z})$ be the usual generator. For $i>k$, choose $u_i$ whose pullback under the quotient map is $x^i$. Naturality of the cup product and the cohomology ring of complex projective space giveTogether with the unit, this determines the ring; equivalently, its reduced part is the ideal $(x^{k+1})/(x^{n+1})$ with the inherited multiplication. This is the cohomology ring of a collapsed projective subspace.

If $n=k+1$, the quotient has just a zero-cell and a $2n$-cell, so it is $S^{2n}$ and is a compact manifold. Conversely, suppose $n\ge k+2$ and the quotient is homotopy equivalent to a compact manifold. Its top cohomology is $\mathbb{Z}$, so that manifold must be closed, orientable, and $2n$-dimensional. But $H^{2n-2}\cong \mathbb{Z}$ while $H^2=0$, contradicting Poincare duality. Therefore

Set $X={\mathbb{CP}}^4/{\mathbb{CP}}^1$ and $A={\mathbb{CP}}^2/{\mathbb{CP}}^1\cong S^4$. Let $u\in H^4(A;\mathbb{Z})$ be a generator and let $a\in H^4(X;\mathbb{Z})$ be the class restricting to $u$. The ring computation in part (a) gives

If $r\circ \iota \simeq {\mathrm{id}}_A$, then homotopy invariance of cohomology gives ${\iota }^{\ast }{r}^{\ast }(u)=u$, hence $r^{\ast }(u)=a$. But $u^2=0$ because $H^8(S^4;\mathbb{Z})=0$, whereas naturality of the cup product would givea contradiction. No such map $r$ exists.