Past exam of the topology and groups course of the University of Oxford / 2023 / 2 / a / iii / Solution

The universal property of a pushout of groups says: for every group $H$ and group homomorphisms $u_j:G_j\to H$ satisfying $u_1{\varphi }_1=u_2{\varphi }_2$, there is exactly one group homomorphism $u:P\to H$ such thatTo prove existence, the universal property of a free product supplies a unique group homomorphism $v:G_1\ast G_2\to H$ with $v{k}_j=u_j$. Every defining relator of $N$ maps toSince $\ker v$ is a normal subgroup, it contains the entire normal closure $N$. The universal property of a quotient group therefore gives a group homomorphism $u:P\to H$ through which $v$ factors. Its restrictions are the required $u_j$.

For uniqueness, $j_1(G_1)$ and $j_2(G_2)$ together generate $P$: their preimages already generate the free product. Consequently a group homomorphism out of $P$ is determined by its restrictions to these two subgroups. Any two candidates for $u$ therefore coincide.

This also proves that the pushout of groups is unique up to a unique compatible group isomorphism. Given another such object $P^{\prime }$, apply the universal property in both directions. Both composites restrict to the original maps from $G_1,G_2$, so uniqueness makes them the respective identity maps.