Epimorphism iff surjective in the category of groups

Statement

The following are equivalent for a homomorphism of groups $\alpha :G\to H$:

1. $\alpha$ is surjective as a set map.
2. $\alpha$ is an epimorphism with respect to the category of groups: For any homomorphisms $\theta _{1},\theta _{2}:H\to K$ to any group $K$, $\theta _{1}\circ \alpha =\theta _{2}\circ \alpha \implies \theta _{1}=\theta _{2}$.

Related facts

• Monomorphism iff injective in the category of groups

Proof

Surjective homomorphism implies epimorphism

This follows simply by thinking of the maps as set maps. In general, for any concrete category, any surjective homomorphism is an epimorphism.

Epimorphism implies surjective homomorphism

The idea here is to define $K$ as the amalgamated free product of two copies of $H$ amalgamated over the image $\alpha (G)$ in $H$, and take $\theta _{1}$ and $\theta _{2}$ as the embeddings of the two copies of $H$ respectively in $K$.

With this approach, we may end up with an infinite $K$ even if $G$ and $H$ are finite. There are slight modifications of this approach that can be used to guarantee that $K$ is finite whenever $H$ is finite. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

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