# Universal Cover of $\Bbb{R}\textrm{P}^2 \vee \Bbb{R}\textrm{P}^2$ in Hatcher

On page 78 of Hatcher here he constructs an example of the universal covering space of $\Bbb{R}P^2 \vee \Bbb{R}P^2$. Now from the picture in example 1.48 I think I understand why it is a covering space. Our covering space map $p$ is the usual quotient map from $S^2 \to S^2/\sim$ when we look "locally" around a point on say the sphere with equator $b^2$. I also get that this space is simply connected because it is just the wedge sum of a bunch of $S^2$'s that has trivial fundamental group by the Van Kampen Theorem.

Now my problem is my understanding of this and the usual construction of the universal cover as consisting of points that are homotopy classes of paths is out of sync. For example, in the picture given in Hatcher I don't get why say the point $ba$ would represent the homotopy class of some path in $\Bbb{R}P^2 \vee \Bbb{R}P^2$. Can someone explain on this a little? We just learned about orbit spaces and covering space actions today and they are a little confusing.

Thanks.

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Recall the theorem which says that the group of deck transformation of the universal cover is isomorphic to the fundamental group of your original space. If you chase through the proof with the example in Hatcher you might get more of an intuition.

Geometrically, you could reason as follows. $\mathbb{R}P^2$ is obtained by taking the quotient of $S^2$ under the antipodal map. So $\mathbb{R}P^2\vee \mathbb{R}P^2$ is the wedge union of two spheres, with each sphere then quotiented by the antipodal map.

Choose the point of identification as a base point. Then a path $a$ which goes halfway round the equator of one of the spheres goes all the way around one copy of $R\mathbb{P}^2$, since antipodal points are identified. Thus it is a loop. Similarly there's a loop $b$ which goes around the other copy of $\mathbb{R}P^2$. These naturally correspond to the $a$ and $b$ in Hatcher's picture.

In particular the deck transformation $ab$ corresponds to the loop going once around one copy of $\mathbb{R}P^2$ and then once around the other copy.

Warning: these kind of geometrical arguments are not always easy. If you aren't careful you can give yourself wrong ideas about a space. I'd recommend trying to get a good feel for how the techniques of covering spaces work abstractly. Once you've derived a result you can then use it to inform your geometrical intuition.

Good luck with learning about covering spaces. They are a bit confusing at first, but they're really cool when you get to know them!

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The universal cover is an infinite string of $S^2$'s not a wedge of just two. (The basepoint of $\mathbb{RP}^2\vee\mathbb{RP}^2$ lifts to two points on each sphere.) –  Grumpy Parsnip Aug 22 '12 at 19:59
I never claimed it was a wedge of just two... I've just explained how the deck transformations of the infinite string of $S^2$s correspond to loops in the wedge union of real projective planes. Is that not clear...? –  Edward Hughes Aug 22 '12 at 20:38
@BenjaLim: no you have to do the identification on each sphere, to get $S^1 / \sim \vee S^1 / \sim$. But I find it helpful to start by imagining $S^1 \vee S^1$ and then doing the identification on each sphere. –  Edward Hughes Aug 22 '12 at 22:11
@BenjaLim: I'm not convinced that your way of looking at it as $S^1 \vee S^1 / \sim$ works. The problem is that then $\sim$ isn't an equivalence relation. My way I don't have to do write down anything mathematically dodgy, I merely imagine the construction in a different way. –  Edward Hughes Aug 22 '12 at 22:15
@hgbreton Yes that sounds more correct. Now I get the bit about you said about deck transformations of the universal cover being isomorphic to the fundamental group of the wedge product. However, I still don't get how the point $ab$ in the universal cover represents a homotopy class of a path in $\Bbb{R}P^2 \vee \Bbb{R}P^2$. –  user38268 Aug 22 '12 at 22:17