We are given group homomorphisms f : H → G and f′ : H → G′, and we wish to find the coproduct. First we have the free product G ∘ G′ with injections i_{G} : G → G ∘ G′ and i_{G′} : G′ → G ∘ G′ such that for group K and pair of homomorphisms j_{G} : G → K, j_{G′} : G′ → K, there exists a unique map h : G ∘ G′ → K such that hi_{G} = j_{G} and hi_{G′} = j_{G′}. This is defined in Lang’s book.

Let P = (G ∘ G′)/N, where N is the normal subgroup generated by elements {i_{G}f(h)i_{G′}f′(h)^{−1}}. This gives a natural commutative diagram:

\begin{matrix}H & ⟶^{f} & G \\ ^{f′}↓ & \overline{i_{G}}↓ \\ G′ & ⟶^{\overline{i_{G′}}} & P \end{matrix}

Now suppose we have another group K and homomorphisms α : G → K and α′ : G′ → K such that αf = α′f′. α and α′ certainly induces β : G ∘ G′ → K such that βi_{G} = α, βi_{G′} = α′. Hence αf = α′f′ ⇒ βi_{G}f = βi_{G′}f′, so that β must factor through P to give \overline{β} : P → K such that \overline{β}\overline{i_{G}} = α and \overline{β}\overline{i_{G′}} = α′.

Next, to show uniqueness, let γ : P → K satisfies γ\overline{i_{G}} = α and γ\overline{i_{G′}} = α′. This gives a composition β : G ∘ G′ → P →^{γ} K, such that βi_{G} = α and βi_{G′} = α′. By property of free product G ∘ G′, β is unique, and hence so is γ.