I am reviewing for an upcoming topology qualifying exam, and I have a question regarding a specific type of question:
Find all connected two-sheeted covering spaces of $S^1\vee \mathbb{R}P^2$ up to equivalence.
The covering space that I was able to think of is picture below.
$p$ sends $a_1$ and $a_2$ to $a$ and $p$ is the anti-podal map on $S^2$ (the universal cover). I have that $\pi_1(\tilde X,\tilde x_0)=\langle a_1,b_1a_2\overline{b_1}|\rangle$ where $\tilde x_0$ is the point where $a_1$ intersects $S^2$, $b_1$ is the half of the equator pictured and $x_0$ is the point at which $a$ intersects $\mathbb{R}P^2$. Then $p_*(\pi_1(\tilde X,\tilde x_0))=\langle a, \gamma a \gamma^{-1}|\gamma^2=1\rangle =\langle a, \gamma a \gamma|\gamma^2=1\rangle$ where $\gamma$ is the generator of $\pi_1(\mathbb{R}P^2)=\mathbb{Z}_2$. Are there other subgroups of index 2 of $\pi_1(S^1\vee \mathbb{R}P^2)=\langle a,\gamma|\gamma^2=1\rangle$? I could try to argue that the only nontrivial covering space of $\mathbb{R}P^2$ is $S^2$, and thus the covering above is all, but this isn't very rigorous. In general, given a space, how might one go about finding all covering spaces (up to equivalence) of a certain index? I know that given a space $X$ (p-conn., locally p-conn., semi-locally simply connected), there is a bijection between the path-connected basepoint preserving covering spaces $p:(\tilde X,\tilde x_0)\rightarrow (X,x_0)$ (up to equivalence) and the subgroups of $\pi_1(X,x_0)$ where the correspondence is given by $p_* \pi_1(\tilde X,\tilde x_0)$. I also know that if $p:(\tilde X,\tilde x_0)\rightarrow (X,x_0)$ is an $n$-sheeted covering, then $p_* \pi_1(\tilde X,\tilde x_0)$ has index $n$ in $\pi_1(X,x_0)$. So, if I can find all subgroups of a given index, then I at least know how many covering spaces I need to find. However, finding all subgroups of a given index on a free group isn't always easy, and it still doesn't tell me how to actually construct these spaces. For example, given my covering space above, How can I prove that I have found all $2$ sheeted covering spaces without calculating all subgroups of a given index? In general, can I find all $n$ sheeted covering spaces without calculating the number of subgroups of index $n$ and just guessing as to the actual construction of the covering space ? Also, as an aside question, what would the universal cover of $S^1\vee \mathbb{R}P^2$ be?
You've found an example which contains a 2-sheeted connected covering of $\mathbb RP^2$. You can stretch that idea to try to imagine other examples: maybe one which contains a 2-sheeted connected covering of $S^1$; maybe one which contains 2-sheeted covering spaces of each of $\mathbb R P^2$ and $S^1$.
Also, finding all index 2 subgroups of a finitely presented group $G$ is easier than you think: every index 2 subgroup of $G$ is normal and has quotient is isomorphic to $\mathbb Z / 2 \mathbb Z$. From this, with a bit more thought, you can deduce that the "kernel" operation induces a bijection between the set of surjective homomorphisms $G \mapsto \mathbb Z / 2 \mathbb Z$ and the set of index 2 subgroups of $G$. So, if you use Van Kampen's theorem to write down a presentation of the fundamental group of your $S^1 \vee \mathbb R P^2$ then you should be able to pretty easily write down a list of all possible surjective homomorphisms to $\mathbb Z / 2 \mathbb Z$.