# 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 $K$, $\theta_1 \circ \alpha = \theta_2 \circ \alpha \implies \theta_1 = \theta_2$.

## 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]