I am reading the book of Massey of Algebraic topology, and I am having trouble to understand this construction.
Let $ S = \left\{ x_i : i\in I \right\}$. For each index $i$, let $S_i$ denote the subset ${x_i}$, and let $F_i$ be an infinite cyclic group $F_i = \{x_i^n : n \in \mathbb{Z}\}$. Let $\phi_i : S_i \to F_i$ denote the inclusion map, $\phi_i(x_i) = x_i^1$. I proved that $F_i$ is a free abelian group on the set $S_i$.
I constructed a homomorphism $f: F_i \to A $ where $A$ is any abelian group and take any function $\psi : S \to A$, with the following rule: $\psi(x_i)^1 = f(x_i^1)$.
Thus, a free abelian group on any set $S$ is a wrak product of a collection of infinite cyclic groups. Because $F$ is tge weak product of the $F_i$, any element $g \in F$ is, for any $i$, the ith component $g_i = x_i^n$ where each $n_i \in \mathbb{Z}$ and $n_i = 0$ for all but finite number of indices $i$. Moreover, $\phi$ is defined by the following rule: $$ (\phi(x_i))(j) = \left\{ \begin{array}{ll} x_i^1 & \textrm{if $i=j$}\\ x_i^0 & \textrm{if $i\neq j$}\\ \end{array} \right. $$ $\phi$ is a one-to-one map, if we wish, we can identify each $x_i \in S$ with its image $\phi(x_i) \in F$. The $S$ becomes a subset of $F$, and it is clear thar we cab express each element $g\neq 1$ of $F$ uniquely in the following form: $g= x_{i_1}^{n_1} x_{i_2}^{n_2} \cdots x_{i_k}^{n_k} $ .
Question 1 : why we can identify $x_i \in S$ with it's image $\phi(x_i)$ of the function $\phi: S\to F$. I think that it is true because the map $\phi_i :S_i \to F_i$ is the inclusion map, $\phi_i (x_i) = x_i^1$. But why it's true for the extension of $\phi_i$?
Question 2: If it's done, $S$ become a subset of $F$, and it is not clear for me that we can express each element $g≠ 1$ of $F$ uniquely in the form : $g= x_{i_1}^{n_1} x_{i_2}^{n_2} \cdots x_{i_k}^{n_k} $.