# Constant Mapping to Identity is Homomorphism/Groups

## Theorem

Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups with identities $e_1$ and $e_2$ respectively.

Let $\phi_e: \struct {G_1, \circ_1} \to \struct {G_2, \circ_2}$ be the constant mapping defined as:

$\forall x \in G_1: \map {\phi_e} x = e_2$

Then $\phi_e$ is a group homomorphism whose image is $\set {e_2}$ and whose kernel is $G_1$.

## Proof

Let $x, y \in G_1$.

Then:

 $\ds \map {\phi_e} {x \circ_1 y}$ $=$ $\ds e_2$ as $x \circ_1 y \in G_1$ $\ds$ $=$ $\ds \map {\phi_e} x \circ_2 \map {\phi_e} y$ as $\map {\phi_e} x = e_2$ and $\map {\phi_e} y = e_2$

So $\phi_e$ is a group homomorphism.

$\blacksquare$