Bijective correspondence between covering space and subgroup of $\pi_1$ My question is quite simple. I've got a lot of trouble to determine all the covering spaces possible of a subspace $X$ given its universal cover $p : \tilde{X} \rightarrow X$. I know there is a bijective correspondence between covering space (up to isomorphism) and subgroup of $G=\pi_1(X)$ (up to conjugacy class). But the link I know is very abstract. Actually, this is what I know :
If we have $H \subset G$ a subgroup of $G$, then it acts on $X$ (because $Aut(p) \cong \pi_1(X)$) this way : if we take $x \in X$ and $\tilde{x} \in p^{-1}(x)$, then $H$ acts on $p^{-1}(x)$ by the monodromy action, and we denote this action by : $\tilde{x} \cdot [\gamma]$. Now, if I consider $y \in \tilde{X}$, I want to know where $y$ is sent, by the action of $[\alpha] \in H$. We know that we have those two bijections :
\begin{array}{rcl}
\phi : \pi_1(X,x) &\to& Aut_{\pi_1(X, x)}(p^{-1}(x))\\
[\gamma] &\mapsto & (\Phi_{[\gamma]} : \tilde{x} \mapsto \tilde{x} \cdot [\gamma]) 
\end{array}
and :
\begin{array}{rcl}
\psi : Aut(p) &\to& Aut_{\pi_1(X, x)}(p^{-1}(x))\\
f &\mapsto & (f_{x} : p^{-1}(x) \rightarrow p^{-1}(x) , \tilde{x} \mapsto f(\tilde{x})) 
\end{array}
So, we have the action :
$$ \pi_1(X, x) \times \tilde{X} \rightarrow \tilde{X} $$
given by :  $$[\alpha] \cdot \tilde{x} =  (\psi^{-1} \circ \phi_{\alpha}) \cdot \tilde{x}$$
(here, $Aut(p)$ is acting on $\tilde{X}$ by the "natural" action : for $f \in Aut(p)$ and $\tilde{x} \in \tilde{X}$, then : $f \cdot \tilde{x} = f(\tilde{x})$), and then if we have a subgroup $H$ of $Aut(p)$ the action is the same but restricted to $H$.
First question : is all this true ? Do I have understand the principle ?
Second question : All of this is very absract, but in practice, how to determine the covering spaces $\tilde{X}/H$ given $\tilde{X}$ the universal covering and $H$ a subgroup of $\pi_1(X)$ ? If we already know what is explicitly $Aut(p)$, it's pretty easy, but if we just know what is $\pi_1$, how to do ? For example, if we take $RP^2 \vee RP^2$, given its universal covering (which is copy of $S^2$ attached on $(1,0,0)$ and $(-1,0,0)$ for each $S^2$, and each $S^2$ is sent one on the first copy of $RP^2$, and the one next to it to the other copy of $RP^2$ and all of this repeated), how to determine all the (connected) covering spaces ? We can take a easier example, for instance $S^1 \vee S^1$, it's the same, I don't see how to do in practice, I will not take all the loop possible of the subgroup I consider, and try to figure out where each points in sent to, right ?
Last question : Linked to the previous question, if I understand well, when we have the universal covers, in order to determine all the connected cover possibles, we should have either a good comprehension of what is $Aut(p)$, or a good comprehension of the $\pi_1(X)$ ? I mean, knowing what $\pi_1$ is like (for example $\pi_1(X) \cong \mathbb{Z}$, but nothing more) is not enough ?
I have some difficulties with this, so sorry if it seems disordered, it's maybe because it is for me.
Thank you for the help ! 
 A: I want to answer your second question first: Lets take your example $X=\Bbb{RP}^2\vee\Bbb{RP}^2$. Using Van Kampen theorem, $\pi_1(X)\simeq \Bbb Z/2\Bbb Z*\Bbb Z/2\Bbb Z$, in particular $\pi_1(X)$ is not finite, so the fiber of the universal covering space can't be finite. The space $Y$ that you gave, which is 
$$Y=\Bbb S^2\amalg \Bbb S^2/_{N_1\sim N_2,~S_1\sim S_2},$$
has a structure of covering space for $X$, but isn't it's universal covering space (Here $N$ and $S$ denote the north and south pole respectively). But $X$ and $Y$ have the same universal covering space, and you can imagine what the universal covering space of $Y$ is; it's
$$\tilde{X}={\amalg_{i\in\Bbb Z}~~ \Bbb S^2_i}/_{N_i\sim S_{i+1}}.$$
(You should do a drawing for this one).
Once this is clear we can try to understand what is $\tilde{X}/H$ for $H=<ab>$ where $\pi_1(X)=<a>*<b>$ for example. For this we need to understand how $ab$ act on $\tilde{X}$. With a good choice for $a$ and $b$, $ab$ sends $\{N_i,S_{i+1}\}$ to $\{N_{i+2},S_{i+3}\}$ (You should really do a drawing here). By uniqueness of covering transformations (two covering transformations are equal if and only if they are equal at one point), $ab$ is just the translation which sends $\Bbb S_i$ to $\Bbb S_{i+2}$ naturally. Hence the quotient space is in fact $Y$, with the same covering space structure as before.
For the first question, I think it is correct. The most important thing you have to understand isn't necessarily that you have an action of $\pi_1$ on the fiber, but that $\psi^{-1}\circ \phi$ is a bijection, and in fact an isomorphism !
For the last question, you understand well. You need to have a good comprehension of what $\psi^{-1}\circ \phi$ is, i.e you have to understand the natural identification of $\pi_1$ and $Aut(\tilde{X})$. In all case (at least when $\pi_1$ is "simple"), you should first determine what the universal covering space and determine $Aut(\tilde{X})$. Then it is easy to quotient by a subgroup of $\pi_1$ because you will just quotient by a group of automorphism.
Edit: I give a picture to see the action of $ab$ on the fiber.

I see $\Bbb{RP}^2$ as a disk with it's boundary quotiented by the action of the antipodal map. The projection of the covering space is the identification with the colors. In blue we have $ab$ and its lift starting at $\tilde{x}_0$.
