# A problem on covering space from Hatcher book...

I was trying a problem from Hatcher's book Algebraic Topology, in section 1.3 problem number 12.

Let $a$ and $b$ be the generators of $\pi_1(S^1 \vee S^1)$ corresponding to the two $S^1$ summands. Draw a picture of the covering space of $S^1 \vee S^1$ corresponding to the normal subgroup generated by $a^2$, $b^2$, and $(ab)^4$ , and prove that this covering space is indeed the correct one.

I know corresponding to every subgroup of this group there exist a covering space. One way to find covering space corresponding subgroup $H$ is to take the orbit space under the action of $H$ on the universal cover. But I am not able to understand how to draw covering space explicitly. Normal subgroup generated by $a^2$ and $b^2$ has infinite index. So clearly covering space will be infinite sheeted and for the second one it will be finite sheeted. I would appreciate any help in this direction.

• You should explain notations, as for those who didn't study hatchers book. Commented Apr 19, 2015 at 1:59
• @BhaskarVashishth These are the common notation used in algebraic topology.I don't think one needs to explain these notation...
– rips
Commented Apr 19, 2015 at 5:25
• You should read this, it is generally bad form to have a title like this.
– user29123
Commented Apr 20, 2015 at 7:14

The covering space of $S^1 \vee S^1$ corresponding to the normal subgroup $N$ is the Cayley graph $\Gamma((\Bbb Z * \Bbb Z)/N)$ of $(\Bbb Z * \Bbb Z)/N$.
That this is a covering space is seen by noting that $\Gamma((\Bbb Z * \Bbb Z)/N)$ is homeomorphic to the orbit space $\Gamma(\Bbb Z* \Bbb Z)/N$, and the covering map $p : \Gamma(\Bbb Z* \Bbb Z)/N \to S^1 \vee S^1$ is given by mapping orbits of $N$-action on $\Gamma(\Bbb Z*\Bbb Z)$ to orbits of $\Bbb Z*\Bbb Z$-action on the graph.
Fact : If $X$ is a path connected, locally path connected and semilocally simply connected space and if a discrete group $G$ acts on $X$ freely and properly discontinuously, then there is a short exact sequence of fundamental groups $$1 \to \pi_1(X) \to \pi_1(X/G) \to G \to 1$$
Since action of $N$ on $\Gamma(\Bbb Z* \Bbb Z)$ is in fact free and properly discontinuous, $\pi_1(\Gamma((\Bbb Z*\Bbb Z)/N)) \cong N$. This proves that $\Gamma((\Bbb Z*\Bbb Z)/N)$ is indeed the covering space corresponding to $N$.