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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 $\mathbb{R}P^2$, 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.)2012-08-22
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    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...?2012-08-22
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    @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.2012-08-22
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    @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.2012-08-22
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    @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 $\Bbb{R}P^2 \vee \Bbb{R}P^2$.2012-08-22
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    @BenjaLim: Hatcher's diagram is misleading. The **point** $ab$ has no real significance. Rather it is the deck transformation which takes the point $e$ to the point $ab$ which corresponds to the homotopy class. Usually this deck transformation is also labelled $ab$. Hatcher has labelled the points in order to make the deck transformations seem obvious.2012-08-22
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    @hgbreton Right I agree with you. In some way I get this action, for example because $ab$ has infinite order in the infinite dihedral group, no point is fixed by the action of this deck transformation which is simply translation by two units. Now yesterday I calculated that the all elements that are not powers of $(ab)$ have order 2. I kinda get that these would correspond to antipodal maps of the sphere, but I don't get the line in Hatcher about the whole string of spheres being flipped.2012-08-22
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    @hgbreton Also, the reason why I said asked how points on the universal cover in the picture correspond to homotopy classes of paths is that I understand once we have a universal cover, given any subgroup $H$ of $\Bbb{Z}/2 \ast \Bbb{Z}/2$ I can construct a covering space $X$ with with fundamental group isomorphic to $H$. This is proposition 1.36 in Hatcher.2012-08-22
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    @JimConant Would you like to post an answer to my question above please?2012-08-22
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    I think the comment about sphere's being flipped is a bit strange. He's just trying to explain that the deck transformations that aren't powers of $(ab)$ take you one sphere to the left/right and then back again. That's clearly illustrated in the diagram by the arrowed line $a$. Ah I see why you were asking about points now. Having read more of Hatcher it seems he likes talking about points. It's different from what I'm used to. That's why I've been talking about deck transformations. I may be unable to clarify exactly about points - sorry!2012-08-22
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    @hgbreton: I misread what you said.2012-08-23