Rotman - Introduction to the theory of groups p.390
Let $\{A_i:i\in I\}$ be a family of groups and let a presentation of $A_i$ be $(X_i|\Delta_i)$, where the sets $\{X_i:i\in I\}$ are pairwise disjoint. Then a presentatioj of $\ast_{i\in I} A_i$ is $(\bigcup X_i|\bigcup \Delta_i)$.
And here is a proof given in the book.
Note that the free product of the free groups $F_i$ with a basis $X_i$ has a basis $\bigcup X_i$.
Let $\pi_i:A_i\hookrightarrow \ast_{i\in I}A_i$ be the imbeddings. If $R$ is the normal closure of $\Delta_i$ and $\phi_i:F_i\rightarrow A_i$ is a surjection with $\ker(\phi)=R_i$, then the map $\Phi:\ast_{i\in I}F_i \rightarrow \ast_{i\in I} A_i$ extending all $F_i\rightarrow A_i \hookrightarrow \ast_{i\in I} A_i$ has kernel the normal subgroup generated by $\bigcup_{i\in I} \Delta_i$.
I completely understand what he said, but think this proof is incomplete. I don't know where he used the fact that "$\bigcup X_i$ is a basis for $\ast_{i\in I} F_i$". Moreover, i think he assumes that $\Phi$ is surjective to conclude his argument, but I don't understand why $\Phi$ is surjective.
Please would someone explain me how to complete this proof.. Thank you in advance
The reason $\Phi$ is surjective is precisely because $\bigcup X_i$ is a basis for $\ast_{i\in I} F_i$: for each $i \in I$, since the map $\phi_i$ takes $X_i$ to a generating set for the group $A_i$, it follows that the map $\Phi$ takes each subset $X_i$ of the union $\bigcup X_i$ to a generating set for the subgroup $A_i$ of the free product $\ast_{i \in I} A_i$. So the entire subgroup $A_i$ of the free product is contained in the image of $\Phi$. But the union of the subgroups $A_i$ generates their free product so $\Phi$ is surjective.