Show that if $A$ is free on a set $S$ in Ab is a coproduct of $|S|$ copies of $\mathbb{Z}$ I am working on the following problem:
Show that if $A$ is free on a set $S$ in Ab, with map $\mathrm{\Phi}:S \rightarrow A$, then $A$ is also a coproduct of $|S|$ copies of $\mathbb{Z}$, where each $\mathrm{\theta}_x :\mathbb{Z} \rightarrow A$ is given by $\mathrm{\theta}_x(n)=n\cdot \mathrm{\Phi}(x)$.
I can't use the fact that $A$ is free on $S$ and the definition of a coproduct to prove that $A$ is isomorphic with the coproduct of $|S|$ copies of $\mathrm{Z}$. How do you construct the diagram? 
I only have to work with the definitions of free object and coproduct.
Definitions:
$X$ is a set. $F$ with $\phi :X \to F$ is free on $X$ in the category $\mathcal{C}$ iff $\forall A \in \mathcal{O}(\mathcal{C})$ and $\psi :X\to A$, $\exists! f \in Mor(F,A)$ such that $f\circ \phi = \psi$.
Suppose $\{A_i : i\in \mathcal{I}\}$ is an indexed family in $\mathcal{O}(\mathcal{C})$. A coproduct of the $A_i$ is an object $A$ together with morphisms $\iota_i \in Mor(A_i,A)$ for all $i\in \mathcal{I}$ satisfying the property: if $B\in \mathcal{O}(\mathcal{C})$, and $\phi_i \in Mor(A_i,B)$ for all $i\in \mathcal{I}$ then $\exists !\tau:A \to B$ such that all the following diagrams are commutative:
$$
\begin{array}[c]{ccc}
A_i&\xrightarrow{\iota_i}&A\\
&\llap{\phi_i}\searrow& \big\downarrow\rlap{\tau}\\
&&B
\end{array}$$
 A: $\begin{array}[c]{ccc}
A&\stackrel{\Phi}{\longleftarrow}&S\\
\Big\uparrow\rlap{\phi_x}&\searrow &\Big\downarrow\rlap{\xi}\\
\mathbb{Z}_x&\stackrel{\iota_x}{\longrightarrow}&\bigoplus\mathbb{Z}_x
\end{array}$
Lets define $\phi_x \in Mor(\mathbb{Z}_x , A)$, so that $\phi_x(n)=n\cdot\Phi(x)$. Since $\bigoplus\mathbb{Z}_x$ is a coproduct, by definition we have $\Rightarrow \exists !f\in Mor(\bigoplus\mathbb{Z}_x,A)$ such that $f\iota_x=\phi_x$. 
Now define $\xi(x)=\iota_x(1)$. Since $A$ is free on $S$, by definition we have $\Rightarrow \exists !g\in Mor(A,\bigoplus\mathbb{Z}_x)$ such that $g\Phi = \xi$.
Lets prove that $A\cong \bigoplus\mathbb{Z}_x$. For this we need that $fg=gf=id_A$:
For all $a\in A$, $\exists x\in S$ such that $\Phi(x)=a$ because $\Phi$ is one-to-one. Hence:
$$fg(a)=fg\Phi(x)=f\xi(x)=f\iota_x(1)=\phi_x(1)=\Phi(x)=a$$
We concluded that $fg=id_A$. Now, for all $z\in \bigoplus\mathbb{Z}_x$, $\exists x\in S$ and $\exists n\in\mathbb{Z}_x$ such that $z=\iota_x(n)$. So:
$$gf(z)=gf\iota_x(n)=g\phi_x(n)=n\cdot g\Phi(x)=n\cdot\xi(x)=n\cdot \iota_x(1)=\iota_x(n)=z$$
So $gf=id_A$ and with this we are done.
