Calculate $p_*(\pi_1(\tilde{X},e_i))$

Question

Give an example of a two-fold cover $(\tilde{X},p)$ of figure eight. For those examples choose a basepoint $e$ and a base point $e_i\in \tilde{X}$ and calculate $p_*(\pi_1(\tilde{X},e_i))$

My attempt:

I have drawn the two examples in the next two pictures. I know that both $(p_i)_*$ are injective maps. I think I read a paper that if in the covering space of figure-eight one edge covers a loop then it is regular.( I might be wrong here). In that logic, the first one should be regular.

One very intuition: in the first picture we have two whole loops of B so this would be one to one map and hence $p_*(\pi_1(\tilde{X},e_i))=2\Bbb Z \star \Bbb Z$ and in the second picture there is no whole loop so in each part there are two 2-1 maps, then $p_*(\pi_1(\tilde{X},e_i))=2\Bbb Z \star 2\Bbb Z$.

I suddenly observe that I wrote the left-hand side pic as fig 1 but forgot to mention that the right-hand side pic is fig 2.

Please help me by guiding how to calculate these problems and if the method is not at least tell me the answers in the comment. Thanks a lot.

Answer 1

EDIT: Full disclosure! The first definition I gave for regular covering spaces is incorrect. (I've starred the incorrect portion.) Although, the covering space that we've constructed using a covering space action is regular. My bad!

A few things to note:

****Incorrect portion:

1. A covering space is regular if and only if it's deck transformation group is normal. (There are other equivalent definitions. You should see Hatcher for more details.)****

The correct portion:

1. A covering space is regular if and only if the action of the deck transformation group is transitive.

You're trying to construct a two-fold cover. Naturally, the cardinality of the deck transformation group of your covering space should be two. Which groups or group has a cardinality of two?

2. I like the first picture a lot! You're on the right track. Consider embedding that triple-8 space into the $x$-$y$ plane of $\mathbb{R}^3$ in the following way:

Note that the embedded space has 180-degree rotational symmetry about the $z$-axis. Thus, $\mathbb{Z}_2$ acts on the triple-8 space by a rotation angle of 180 degrees about this axis. That is, the action of $\mathbb{Z}_2$ identifies two points on the triple-8 space if and only if they're related by a 180 degree rotation.

In fact, this action of a group on a topological space describes a covering space action. See Hatcher again to fill in the details. The fundamental group of the covering space given by this covering space action is precisely $2\mathbb{Z} \star \mathbb{Z}$---just as you calculated. You just need to argue that the covering space that me and you have constructed is a covering space by chasing the definitions in Hatcher (i.e. look up covering space action).