# Irregular covering space of $\mathbb{R}P^2\vee\mathbb{R}P^2$

This was on my final last semester (to find such a cover), and I missed it. Here are my thoughts on it since then:

I know that the universal cover of $X = \mathbb{R}P^2\vee\mathbb{R}P^2$ is (loosely) a string of 2-spheres that goes off in both directions.

I also know that the fundamental group of $X$ is the group with presentation $<a,b|a^2=b^2=1>$. Any covering space of $X$ will have a fundamental group that is (isomorphically speaking) a subgroup of this group. Furthermore an irregular covering space will have fundamental group that is isomorphic to a non-normal subgroup of $\pi_1(X)$. $\mathbb{Z}_2$ is such a group.

Now I want to find a covering space of $X$ that has fundamental group $\mathbb{Z}_2$. This is where things get a little shaky for me. How do I find such a space. I know that the number of sheets will be infinite since the number of cosets of a finite subgroup in a infinite group is infinite.

So here's what I'm thinking. If I take my infinite string of 2-spheres and quotient one of them into an $\mathbb{R}P^2$ via the antipodal map (getting something like an $\mathbb{R}P^2$ with two strings of 2-spheres coming off a single point). I think this will do the trick. I know this is being a little sloppy (not explicitly giving the maps, but I hope it is obvious- If not I can clarify).

So I guess what I'm hoping is that someone can help me figure out if this is right or wrong, but even more what I'm hoping for is that someone can share some insight into this problem (or this sort of problem in general)- maybe provide a better way to look at this problem.

Sorry if this was long winded.

Thanks much :)

Edit: I believe solution is also wrong. For any neighborhood of a point that maps to the point that is the wedge of $\mathbb{R}P^2$ and 2 $S^2$'s will not be homeomorphic to any neighborhood of the point where the two projective planes are joined. I believe Dylan has a correct solution, but I am still hoping for some insight.

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@Dylan I believe you are correct. Now that I think about it more. I don't think the space I described is even a covering space. I'm still hoping to gain some insight into figuring out this sort of problem if anyone can help. –  MJoszef Jul 8 '11 at 4:12
A similar question is to determine a covering space that has fundamental group $\mathbb Z$. Although it's probably a good deal easier. –  JSchlather Jul 8 '11 at 7:23

I believe that this is $\mathbf{R}P^2$ with an infinite (in one direction) string of spheres attached to it. I can think of a couple ways to arrive at this. Here's one, which requires knowledge of the universal cover:
Let $U$ be the universal cover of $X$, which as you say is an infinite (in two directions) string of spheres. If $Y$ is a cover of $X$ with fundamental group $\mathbf{Z}/2\mathbf{Z}$, then $U$ is also the universal cover of $Y$. The covering $U \to Y$ has degree $2$ and is normal/regular, so there is some subgroup, isomorphic to $\mathbf{Z}/2\mathbf{Z}$, of the deck transformations of $U \to X$ such that $U/G \approx Y$. The obvious thing to try (and I think these are the only elements of order two) is the deck transformation which is the antipodal map on one fixed sphere, and which flips the rest of the chain around that sphere. I think this quotient works.
I hesitate to write about the alternative I have in mind, since it feels less justifiable, but maybe it will help: Since a covering of $X$ restricts to a covering of $\mathbf{R}P^2$, you know that you must stick a bunch of spheres and projective planes together. Sometimes it helps me to draw some dots along a vertical line, representing the points of the cover lying above the point where the $\mathbf{R}P^2$ meet; on the left side are the things mapping to the first copy of $\mathbf{R}P^2$. Then on either side of a particular dot I can place a copy of $\mathbf{R}P^2$ or half of a sphere, which must be made whole at another point. If you play around with this for a while, I think it should yield the answer as well.