= Solution
Use the standard integral <cohomology ring> $H^*(\mathbb{CP}^2)=\mathbb Z[u]/(u^3)$, $|u|=2$. Any self-homeomorphism $h$ sends $u$ to $u$ or $-u$, and hence sends $u^2$ to $u^2$. Its <Lefschetz number> is therefore $1\pm1+1$, which is either $3$ or $1$. By the <Lefschetz fixed-point theorem>, every such <homeomorphism> has a fixed point.
Now $\mathbb{CP}^2$ is <simply connected>: its CW structure has no one-cells and only two- and four-cells. If it covered a connected space with degree $d\ge2$, it would be the <universal cover>. The usual <covering space> hypotheses hold for the base because it is locally homeomorphic to this manifold. Its <deck transformation group> would act transitively on every fibre and have $d$ elements. A nonidentity <deck transformation> has no fixed point: uniqueness of lifts makes one fixing a point the identity everywhere on the connected cover. This contradicts the preceding calculation. Thus
$$
\boxed{\mathbb{CP}^2\text{ is not the total space of a covering of degree at least two}.}
$$
Thus the <complex projective plane has no nontrivial covering quotient>.
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