Why the plane isn't universal covering space of $\mathbb RP^2$?

Question

We've recently learnt about covering spaces in my university topology class , and universal covering spaces. I'm finding it hard to understand why for example the mobius strip and the Klein bottle both have the plane as a universal covering space but $\mathbb {RP^2}$ has the sphere.

I'm aware that $\mathbb {RP^2}$ can be defined as the antipodal points on $S^2$ but also that it is the quotient of the square, the same as the Mobius strip and Klein bottle, so why wouldnot it have the same covering space as them?

Answer 1

As we know that the fundamental group of $\mathbb RP^2= \mathbb Z_2$ (I guess you are familiar with Van-Kampen and CW-structure of real projective plane). So if $\mathbb R^2$ is a universal cover, then it will be a $2-sheeted$ cover of projective plane. Now just using the definition of covering you can prove that universal cover of projective palne is compact (WHY?? just try to prove that every cover has a finite sub-cover, and the fact that real projectiv space is compact). Thus plane cannot be its universal cover.

Answer 2

Here is why the plane $\mathbb R^2$ is not a covering of $\mathbb {RP}^2$, and thus even less a universal covering. Consider the univeral covering map $$p:S^2\to \mathbb {RP}^2:(x,y,z)\mapsto [x,y,z]=(x,y,z)/\pm Id$$ If there existed a covering map $f:\mathbb R^2\to \mathbb {RP}^2$, by simple connectedness of $S^2$ the map $p$ could be lifted to a covering map $P:S^2\to \mathbb R^2$ satisfying $p=f\circ P$.
But then the image $P(S^2)\subset \mathbb R^2$ would be bounded by compacity of $S^2$ and thus $P$ would certainly not be surjective.
This however is a contradiction: the covering map $P:S^2\to \mathbb R^2$ must, like all covering maps with connected codomain, be surjective.

Answer 3

Apologies for the lack of imagery (as this is fundamentally a geometric question), I'll try to explain the picture. I feel the other answers do not give you the whole picture and cover what is confusing you, such as what you've commented on, so I feel an additional answer may help you (albeit late).

There are many ways to see that $\mathbb{R}^2$ is not a universal cover from a high-level point of view, and I think that this is what the other answers cover. For example, as the universal cover is unique, as we know it to be $S^2$, and as $S^2$ is not homeomorphic to $\mathbb{R}^2$, we have that $\mathbb{R}^2$ is not the universal cover. Alternatively, the fundamental group of $\mathbb{RP}^2$ is $\mathbb{Z}_2$, so the preimage of a point in $\mathbb{RP}^2$ must have two representatives in the universal cover, and this will not be the case in your tiling. However, I will also show this from a low-level point of view, as I expect this is where your confusion lies.

Draw out what you believe to make the universal cover of $\mathbb{RP}^2$; a tiling of the plane into squares, labelling the sides of the squares with how (what you believe to be) the covering map $p:\mathbb{R}^2 \to \mathbb{RP}^2$ sends sides of the square into $\mathbb{RP}^2$ (i.e. with single and double arrows).

Consider a vertex $\bar{x}_{0}$ of one of the squares in your tiling, and consider $p(\bar{x}_0)$, a point in $\mathbb{RP}^2$. Then, as your $p$ is a covering map, there exists an open set $U \subset \mathbb{RP}^2$ about $x$ such that $p^{-1}(U)$ is a disjoint union of open sets, each of which is homeomorphic to $U$ by $p$.

So take such a $U$, and draw onto your tiling its preimage in $p$. It should look like the unions of sets around half of the vertices in your tiling, each of which is homeomorphic via $p$ to $U$. However, considering just one of these (call it $U_{p^{-1}}$), consider a point of it on one of the sides of the square, say this is on the vertical line, slightly north of the vertex. Then it also contains another point on the vertical line but slightly south of the vertex, both of which are mapped to the same place (considering the equivalence relation defined on the edges of the squares).

So $p$ maps both points to the same place, and therefore is not injective, and thus not a homeomorphism. Therefore, $p$ is not a covering map.

With both the torus and the Klein bottle, you do not run into this problem.