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?

  • $\begingroup$ I'm not sure off hand that the plane even is a covering space of the projective plane. My gut says 'no,' since the universal covering space is universal - there should only be one up to homeomorphism. $\endgroup$ Nov 24, 2016 at 16:50
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    $\begingroup$ @AlfredYerger If the plane were a covering space of the projective plane, the sphere would cover the plane. But the sphere is compact. $\endgroup$ Nov 24, 2016 at 16:58
  • $\begingroup$ Can we use the face that $\pi_1(\mathbb{RP^2})=\mathbb Z_2$? $\endgroup$ Nov 24, 2016 at 17:02
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    $\begingroup$ Why would it be? $\endgroup$ Nov 24, 2016 at 17:17
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    $\begingroup$ @thatguy : the projective plane is a quotient of $\Bbb R^2 \setminus \{(0,0)\}$, not of $\Bbb R^2$. $\endgroup$
    – Watson
    May 25, 2017 at 14:40

3 Answers 3


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.

  • $\begingroup$ Ok thank you that makes sense, perhaps it is my understanding of the universal covers for the Mobius strip and Klein bottle that was wrong then, I had thought they were the plane due to the fact we can tile the plane with the square then have a covering map from this tiling to the square, if this isnt the reason the plane is the universal covering of these then how would you go about showing that the plane is the universal covering? $\endgroup$
    – thatguy
    Nov 24, 2016 at 17:49
  • $\begingroup$ If the plane is any covering of the space $X$ then the plane is the universal covering of $X$, that's all. The reason is that the plane is simply connected. $\endgroup$ Nov 24, 2016 at 18:46

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.


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.

  • $\begingroup$ Thanks this makes sense as well, can I ask you the same question I left on Georges response then? $\endgroup$
    – thatguy
    Nov 24, 2016 at 18:12
  • $\begingroup$ @thatguy What question?? $\endgroup$ Nov 25, 2016 at 10:58
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    $\begingroup$ I think my understanding is wrong about how the plane is then a covering of the Mobius strip and Klein bottle, I thought it was due to their quotient of the square and the fact we can tile the plane using the square, if this isnt the reason then how can we show it is a covering space? I don't really understand Hatchers use of the fundamental group for this $\endgroup$
    – thatguy
    Nov 26, 2016 at 14:20
  • $\begingroup$ I didn't get your doubt, but I think one way to prove the covering space is explicitly defining some covering map. $\endgroup$ Nov 26, 2016 at 14:28

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