Let $X$ be a path-connected, locally path-connected and semilocally simply-connected topological space. (Let $(Y,p)$ be a covering space of $X$, then it is called an abelian covering space if it is normal and its automorphism group as a $X$-covering space is abelian. It's called a universal abelian covering space if it is abelian and for any abelian covering space $(Z,q)$ of $X$, there is division $r$, such that $p =rq$.) Prove that $X$ has a universal abelian covering space, and it's unique up to isomorphism. Construct an abelian universal covering space for $S^1 \vee S^1$.
Since $X$ is path-connected, locally path-connected and semilocally simply-connected, it has a universal covering space. Does the existence of a universal covering space have something to do with the existence of a universal abelian covering space? Moreover, the uniqueness part does not seem to be easy for me... Let $(Y_1,p_1)$ and $(Y_2,p_2)$ be two universal abelian coverings of $X$, then by definition, there exist $r_1:: Y_1 \rightarrow Y_2$ and $r_2:Y_2 \rightarrow Y_1$ such that $p_1 = p_2 \cdot r_1$ and $p_2 = p_1 \cdot r_2$. Then are $r_1$ and $r_2$ isomorphisms and inverses to each other?
Thanks very much.