Prove Theorem 68.4 (uniqueness of free products): let\(\ts{G_\alpha}_{\alpha \in J}\)be a family of groups. Suppose\(G\)and\(G'\)are groups and\(i_\alpha: G_\alpha \to G\)and\(i_\alpha': G_\alpha \to G'\)are families of monomorphisms such that the families\(\ts{i_\alpha(G_\alpha)}\)and\(\ts{i_\alpha'(G_\alpha)}\)are free products with the extension property. Then there is a unique isomorphism\(\phi: G \to G'\)such that\(\phi \circ i_\alpha = i_\alpha'\)for each\(\alpha\).
Solution
By the extension property for the free product family \(\{i_\alpha(G_\alpha)\}\subset G\), applied to the homomorphisms \[ i'_\alpha:G_\alpha\to G', \] there is a unique homomorphism \[ \phi:G\to G' \] such that \(\phi\circ i_\alpha=i'_\alpha\) for every \(\alpha\).
Likewise, the extension property for \(G'\) gives a unique homomorphism \[ \psi:G'\to G \] such that \(\psi\circ i'_\alpha=i_\alpha\) for every \(\alpha\).
For each \(\alpha\), \[ (\psi\phi)\circ i_\alpha=\psi\circ i'_\alpha=i_\alpha. \] The identity map \(1_G\) has the same property. By uniqueness in the extension property for \(G\), \[ \psi\phi=1_G. \] Similarly, \[ \phi\psi=1_{G'}. \] Hence \(\phi\) is an isomorphism with inverse \(\psi\).
If \(\phi':G\to G'\) is another isomorphism satisfying \(\phi' i_\alpha=i'_\alpha\) for every \(\alpha\), then the uniqueness clause in the extension property for \(G\) gives \(\phi'=\phi\). Thus the required isomorphism is unique.