I'm working on this problem: Find all connected $3$-sheeted covering space of the wedge sum of a circle and the projective plane.

Here is my work:


Let me explain: So we start with the preimage of the common point in the wedge sum, which is 3 points. Now in case 1, I suppose the preimage of the projective plane is 3 copies of it. Then each copy is linked to each point. To make this connected, I need a "circle" connecting each 2 points, hence the two lines I drew. I only have 1 line left to add (because this is a 3-sheeted covering), but I need at least 2 more lines to make this a covering, hence this is impossible.

In case 2, I suppose the preimage of the projective plane is a copy of it and its 2-sheeted covering space, the sphere $S^2$. Then $S^2$ must connect 2 points and the copy of projective plane has nowhere to go but to link with the remaining point. I need the covering to be connected, hence I drew 2 "circles" to connect the isolated points to others. I have one line left to add, and the lowest point in my work need to connect with a circle, so I put it there.

To sum up, there is only one 3-sheeted cover?

My question: is this correct? I don't know the optimal way to consider all cases so any other methods will be great.

  • $\begingroup$ Are you familiar with covering spaces being in correspondence with subgroups of the fundamental group? $\endgroup$ – Osama Ghani Aug 12 '20 at 10:53
  • $\begingroup$ Also as a hint, your first one almost works! You only need one more line, see if you can find it :) $\endgroup$ – Osama Ghani Aug 12 '20 at 12:21
  • $\begingroup$ I know the theorem but not familiar with the construction $\endgroup$ – T C Aug 12 '20 at 14:06

One of the best way to construct coverings of wedges is to take wedges of coverings, which is definitely the approach you took! Here's what you forgot about the first one. $S^1$ is itself a 3 sheeted covering of itself (take $S^1$ and trisect it). Now at the three trisection points on $S^1$, you can wedge an $\mathbb{R}P^2$ on. This gives a three sheeted cover. Although visually it may be "obvious", you can check. If we choose one of the lifts of the wedge point as a base point, then the subgroup is generated by $\langle a^3, b, aba^{-1}, a^2ba^{-2} \rangle$, where $a$ is the generator for $S^1$ and $b$ is the generator for $\mathbb{R}P^2$. Then the quotient group is just given by $\langle a|a^3 \rangle \cong \mathbb{Z}/3$ so we have shown it's a three sheeted covering. This was close to your first attempt, all you had to do was join the top and bottom by an edge.

I don't think your second one cannot work. If you lift the generator of the fundamental group of the $\mathbb{R}P^2$ factor, then it generates the $\mathbb{R}P^2$ factor upstairs, but is nullhomotopic in the $S^2$ factor. This cannot happen since choosing a different lift point just corresponds to a different conjugacy class and so you can't have a lift to $0$ using one basepoint and a lift to something non-zero at another basepoint.

  • $\begingroup$ the thing I hate the most about AT is that I have no way to verify the answer. Can you confirm to me that there is only 1 cover? $\endgroup$ – T C Aug 12 '20 at 16:10
  • $\begingroup$ It's definitely the only normal/regular covering. You can do some quick algebra to check that. As for whether there are any non-normal coverings, I'm not sure. $\endgroup$ – Osama Ghani Aug 12 '20 at 19:49

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