I am trying to solve the following exercise.
Let $a$ and $b$ be the generators of the fundamental group $\pi_1(\mathbb R\mathbb P^2\vee \mathbb R\mathbb P^2)$ and consider the subgroups $H=\langle (ab)^n\rangle$ and $K=\langle (ab)^n,a\rangle$. Find the covering spaces $p_H:X_H\to \mathbb R\mathbb P^2\vee \mathbb R\mathbb P^2$ and $p_K:X_K\to \mathbb R\mathbb P^2\vee \mathbb R\mathbb P^2$ corresponding to the subgroups $H$ and $K$ respectively.
There are very nice post in this website (and some PDFs in internet) that gives the solution of this exercise. I need to convince myself that the solutions given are indeed the correct one.
Let us start with $p_H$. Consider $X_H$ a string $2n$ spheres $S^2$ wedged together so that each sphere are wedge to other spheres at both the north and south poles. The following figure shows the case $n=2$:
Let $G$ be the presentation group $\langle a,b\,|\, a^2,b^2, (ab)^n\rangle$. It is easy to see that $G$ has only four elements: the identity $e$, $a$, $b$ and $ab$. Then we can define an action $p:G\to \textrm{Hom}(X_H)$ of $G$ over $X_H$ by a natural manner so that the orbit space $X_H/G$ is homeomorphic to $\mathbb R\mathbb P^2\vee \mathbb R\mathbb P^2$. By the Propositiion 1.40 of Hatcher's book, we can conclude that $G$ must be isomorphic to: $$\frac{\pi_1(\mathbb R\mathbb P^2\vee \mathbb R\mathbb P^2)}{(p_H)_\star(\pi_1(X_H))}$$ This implies that $(p_H)_\star(\pi_1(X_H))$ is isomorphic to $H$.
However in the case of the subgroup $K$ I cannot procedure in a similar way. How the action group $G$ should be defined over some space $X_K$?
