I saw the following conclusion from a reference book:
Let $G=*_{i\in I}G_i$ be free product of $G_i$, where each $G_i$ is a group.
If $I=\{1,\cdots, n\}$, each $G_i=\Bbb Z$, then $G$ is isomorphic to the free group $F_n$.
My question is : since each element in $G$ is not uniquely expressed by elements of $G_i$, how to constfuct the explicit bijective map?
$\renewcommand{\phi}{\varphi}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$Use the universal properties. Let $G_{i} = \Span{g_{i}}$ be infinite cyclic, and let $F_{n}$ be the free group on the symbols $h_{i}$.
By the universal property of the free groups, there is a unique homomorphism $\phi : F_{n} \to G$ such that $\phi(h_{i}) = g_{i}$ for all $i$.
The maps $\psi_{i} : G_{i} \to F_{n}$ given by $\psi(g_{i}^{k}) = h_{i}^{k}$ are clearly group homomorphisms, as $G_{i}$ is infinite cyclic. By the universal property of free products, there is a unique homomorphism $\psi : *_{i} G_{i} \to F_{n}$ such that $\psi(g_{i}) = h_{i}$ for each $i$.
Now $\phi$ and $\psi$ are inverse to each other.