Prove or disprove that $\mathbb{S}^4$ is a covering space of $\mathbb{C}\text{P}^2$

Solution 1:

It is well-known that $\mathbb CP^2$ has a CW-structure with one cell in dimensions $0,2,4$. Hence the fundamental group of $\mathbb CP^2$ agrees with the fundamental group of the $2$-skeleton which is a copy of $S^2$. Thus $\mathbb CP^2$ is simply connected.

Since $\mathbb CP^2$ is simply connected, it is its own universal covering space. Universal covering spaces (i.e. simply connected covering spaces) are unique up to isomorphism of covering spaces. Thus if there exists a covering space $p : S^4 \to \mathbb CP^2$, we must have $S^4 \approx \mathbb CP^2$.

The homology groups $H_k(\mathbb CP^2)$ are $\mathbb Z$ for $k = 0,2,4$ and $0$ else. This follows easily by computing the cellular homology groups. See also Representatives of generators for the homology group of the complex projective space. Therefore $S^4 \not\approx \mathbb CP^2$.