Is the Universal Covering of $S^1 \vee S^2$ as a covering space unique? In the question Universal Covering of $S^1\vee S^2$ there is a suggestion of a universal covering, namely the real line with spheres attached to every integer.
I actually understand, why is that the case. But I am having a bit trouble to understand that this is simply connected, even though intuitively makes sense.
Furthermore, I would like to know if there is any other covering space (up to equivalence). Actually, since $π_1(S^1 \vee S^2) = \mathbb{Z}$ and since this space is enough good for the equivalence of subgroups and coverings spaces to be true. I am thinking that maybe there is only one covering space, since every non-trivial subgroup of $\mathbb{Z}$ is isomorphic to $\mathbb{Z}$. Pretty much like the case of circle which up to equivalence there is only the circle and the line as covering space.
 A: First, you can indeed use Van Kampen's Theorem to show that the real line with spheres attached at every integer is simply connected: first show by induction on $n$ that it is true with spheres attached to only $n$ integers; and then do a direct limit argument.
Let me explain (briefly?) how covering space theory applies to answer your followup question. By the way, a good book for mastering covering space theory is Munkres "Topology".
For covering space theory to work well one needs to limit one's attention to topological spaces that are path connected, locally path connected, and semilocally simply connected. I will make those assumptions tacitly in what I write. Your example $S^1 \vee S^2$ is a wedge of two manifolds; every manifold satisfies those properties (trivially true); and every wedge of (nice enough) spaces that satisfy those properties also satisfies those properties.
The preliminary theorem is that if $f : X \to Y$ is a covering map between path connected spaces, and if $p \in X$ and $q = f(p) \in Y$, then the induced homomorphism of fundamental groups
$$f_* : \pi_1(X,p) \to \pi_1(Y,q)$
$$
is injective, and so it induces an isomorphism from $\pi_1(X,p)$ to $\text{image}(f_*)$.
Consider now two covering maps $f_1 : X_1 \to Y$ and $f_2 : X_2 \to Y$.
To say that these two maps are equivalent means that there exists a homeomorphism $h : X_1 \to X_2$ such that $f_1(X_1)=X_2$.
The main classification theorem has two parts:

$f_1$ is equivalent to $f_2$ if and only if the two subgroups $\text{image}((f_1)_*)$ and $\text{image}((f_2)_*) < \pi_1(Y,q)$ are conjugate subgroups of $\pi_1(Y,q)$.


For every subgroup $H < \pi_1(Y,q)$ there exists a covering map $f : X \to Y$, and points $p \in X$ and $q=f(p) \in Y$, such that $\text{image}(f_*) = H$.

There are several variations on this theorem (e.g. classification up to base-point-preserving covering equivalence; classification of regular covering maps), but this one will suffice for your question.
Now, the wording of your question contains a mis-understanding: one does not really classify covering spaces; instead one classifies covering maps. Also, it does not matter whether two subgroups are abstractly isomorphic: what matters is whether they are conjugate. (And for the base-point-preserving version of the theory, what matters is whether the two subgroups are the same subgroup).
So, starting with $S^1 \vee S^2$, first use Van Kampen's theorem to compute that its fundamental group is infinite cyclic, generated by a loop that goes once around the $S^1$.
The infinite cyclic group $\mathbb Z$ has many subgroups, namely $\mathbb Z$ itself, and the trivial group, and the infinite cyclic subgroups $n \mathbb Z$ for each $n \ge 1$. Since $\mathbb Z$ is abelian, each of these subgroups is normal, hence it is the only subgroup in its conjugacy class, hence the covering maps are on one-to-one correspondence with the subgroups. And here are the domains of all of those covering maps, i.e. all of the different covering spaces:

The covering map corresponding to the trivial subgroup is the universal covering space (this is always true), which in this case is the real line with spheres attached to each intger point.


The covering map corresopnding to $n \mathbb Z$ is the circle with spheres attached to $n$ different points on that circle. So for the case $n=1$, the covering space corresponding to the whole subgroup $\mathbb Z$ itself is just $S^1 \vee S^2$ itself.

