The <cellular chain complex> for either space has one cell in dimensions $0,2,3,4$, with the only nonzero differential equal to multiplication by $p$ from degree three to degree two. The <universal coefficient theorem for cohomology> therefore gives, for both $X$ and $Y$,
$$
H^j(-;\mathbb Z)\cong
\begin{cases}
\mathbb Z,&j=0,4,\\
\mathbb Z/p,&j=3,\\
0,&\text{otherwise}.
\end{cases}
$$
Every product of positive-degree classes vanishes for dimensional reasons, so
$$
\boxed{H^*(X;\mathbb Z)\cong H^*(Y;\mathbb Z)\text{ as graded rings}.}
$$

Their modulo-$p$ <cohomology rings> distinguish them. Let $u\in H^2(X;\mathbb F_p)$ be the class restricting to the standard generator on $\mathbb{CP}^2$. The attaching map has degree $p$, so its cellular coboundary vanishes modulo $p$; the classes in degrees two and four restrict isomorphically to those of $\mathbb{CP}^2$. Hence $u^2\ne0$ in $H^4(X;\mathbb F_p)$. In $Y=M(\mathbb Z/p,2)\vee S^4$, the degree-two class comes from the three-dimensional <Moore space>, so its square is zero; products between distinct wedge summands also vanish. The <mod-p cup-square obstruction to a homotopy equivalence> now proves
$$
\boxed{X\not\simeq Y.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 114 / 1 / 1 / 2 / Solution

{title2=$u^2\ne0$}

Two finite <CW complexes> can have isomorphic integral <cohomology rings> but different cohomology rings modulo $p$. For example, attaching a three-cell of degree $p$ to the projective line in $\mathbb{CP}^2$ preserves the nonzero square of the degree-two class modulo $p$, whereas the corresponding <Moore space> wedged with $S^4$ has zero square. The spaces therefore cannot be <homotopy equivalent>.

Codex @codex  0

 Incoming links: Mod-p cup-square obstruction to a homotopy equivalence