Covering of a space given by epimorphisms from the fundamental group. Let $X$ be a connected finite CW-complex. Let $A$ be a finitely generated abelian group. I read the following:

Any epimorphism  $f:\pi_1(X)\longrightarrow A$ gives rise to a connected regular covering $p_f:X^f\longrightarrow X$ with group of Deck transformations $A$.
Conversely, any connected regular $A-$cover $p:Y\rightarrow X$ yields
$$1\rightarrow \pi_1(Y)\stackrel{p_*}{\rightarrow} \pi_1(X)\stackrel{f}{\rightarrow} A\rightarrow 1$$
so that there is an $A-$equivariant homeomorphism $Y\cong X^f$.

Could you please help me understand this construction. I'm aware that connected coverings of $X$ are given up to isomorphism by subgroups of $\pi_1(X)$ but I don't see the connection with this epimorphism.
 A: This is an immediate consequence of the theory of regular covering maps.
As you know, a connected covering map $p : U \to X$ taking a base point $u \in U$ to a base point $x \in X$ induces an injection $p_* : \pi_1(U,u) \to \pi_1(X,x)$, and the subgroup of $\pi_1(X,x)$ that corresponds to $f$ is just $\text{image}(p_*)$. 
So, what you also need to know, from the theory of regular covering maps, is that $\text{image}(p_*)$ is a normal subgroup of $\pi_1(X,x)$ if and only if $p : U \to X$ is a regular covering map, which means that its deck transformation group acts transitively on the fiber $p^{-1}(x)$. And furthermore, if this is the case, then the deck transformation group is isomorphic to the quotient group $\pi_1(X,x) / \text{image}(p_*)$. Putting this all together, the regular covering maps are classified by the normal subgroups.
The connection with an epimorphism comes from group theory: for any epimorphism $f : \pi_1(X,x) \to A$, if we let $N = \text{kernel}(f)$ then $A$ is isomorphic to the quotient group $\pi_1(X,x) / N$. And then we use regular covering theory to construct the covering map $p : X_f \to X$ corresponding to the normal subgroup $N$.
