Relationship between the fundamental group and the natural equivalence classes of its universal cover For a universal covering $p: Y \to X$, under the equivalence relation $y_1 \sim y_2$ if $p(y_1) = p(y_2)$, $Y$ admits the quotient map $\, \, \, q: Y \to Y / \sim$. There is a natural bijection $\bar p : Y / \sim \, \,  \to X$ and according to a few wikipedia examples there is an isomorphic relationship between the fundamental group of $X$ and the equivalence classes. One such example from wikipedia is 

The universal cover of a circle $S^1$ is $\mathbb{R}$ and we have $S^1
 = \mathbb{R} / \mathbb{Z}$. Thus $\pi_1(S^1, x) = \mathbb{Z}$ for any base point $x$.

Is this obvious as wikipedia seems to suggest it is, if so can someone please explain why? If not could someone please prove this? 
EDIT: From the answers it seems my exact questions is not clear; I understand the concept of a universal cover and the fact it comes down to finding a simply connecting covering. My questions is focusing on why, in the example, $\pi_1(S^1, x) = \mathbb{Z}$ ? Does this generalise ot the equivalence relation  I've defined in the first paragraph? 
 A: It's not trivial, just known. If $p:E\rightarrow B$ is a covering map and E is path connected, $b\in B, e\in p^{-1}(b) $ then we can define $F:\pi_1(B,b)\rightarrow p^{-1}(b)$ by $F([\varphi])=\hat \varphi ^e(1)$, where $\hat \varphi ^e(1)$ is the unique path in E which is the lift of $\varphi$ and starts at $e$. It's well defined since if two paths have a base-point preserving homotopy, then their lifts end at the same point. Also, F is surjective, and if $E$ is also simply connected, then F is bijective. Now, in our case of $p:\mathbb{R} \rightarrow S^1 , p(t)=e^{2\pi i t}, p^{-1}(1)=\mathbb{Z}$ F is also a group homomorphism and hence an isomoprhism. Notice that it only applies to the circle - we don't generalize this, but use different means based on this result (e.g. Van Kampen).
A: Finding the universal cover of $X$ just comes down to finding a simply connected space which covers $X$. One of the prototypical examples of a covering map is the map $p\colon \mathbb{R}\rightarrow S^1$ given by the exponential function, and given that $\mathbb{R}$ is a contractible space, it is certainly simply connected. It follows that $\mathbb{R}$ is a covering space for the circle $S^1$.
You might like the try and convince yourself that the sphere $S^2$ is its own universal cover. Just remember the definition of a universal cover.
A: Well, it is trivial if you already know the answer.
But, perhaps, what you wish to know is about "covering action" of groups.
If you have an action satisfying some properties, you have a universal covering corresponding to this action. It is the case of $\mathbb{Z}$ acting in the real line and it is the case of $\mathbb{Z}_2 $ acting on the sphere (and you getting the projective plane).
You need some conditions... I assume that you know what is an action of a group.
Let $G$ be a group and $X$ be a topological space.
Assume that $G$ acts on $X$ such that $p: X\to X/G$ is a covering. Then you have that $ Aut (p) = G $. And, more, $p$ is a regular/normal covering.
In the case of circumference, we have that $\mathbb{Z}$ acts in $\mathbb{R}$ such that $p:\mathbb{R}\to\mathbb{R}/\mathbb{Z} $ is a covering map. Then you can conclude that $Aut(p)\cong \mathbb{Z}$. But, since $p$ is a universal convering, $\pi _1(\mathbb{R}/\mathbb{Z}, \ast )\cong Aut(p)$.
Same for the sphere!
