# Mapping to Identity is Unique Constant Homomorphism

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## Theorem

Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups whose identities are $e_G$ and $e_H$ respectively.

Then there exists a unique constant mapping from $G$ to $H$ which is a homomorphism:

- $\phi_{e_H}: G \to H: \forall g \in G: \map {\phi_{e_H} } g = e_H$

## Proof

Let $h \in H$ such that $\phi_h: G \to H$ is a (group) homomorphism, where $\phi_h$ is defined as:

- $\forall g \in G: \map {\phi_h} g = h$

Then from Group Homomorphism Preserves Identity:

- $\map {\phi_h} {e_G} = e_H$

and so $h = e_H$.

Hence the result by definition of constant mapping.

It remains to prove that such a constant mapping is in fact a homomorphism.

Let $x, y \in G$.

\(\ds \map {\phi_{e_H} } {x \circ y}\) | \(=\) | \(\ds e_H\) | Definition of $\phi_{e_H}$ | |||||||||||

\(\ds \) | \(=\) | \(\ds e_H * e_H\) | Definition of Identity Element | |||||||||||

\(\ds \) | \(=\) | \(\ds \map {\phi_{e_H} } x \circ \map {\phi_{e_H} } y\) | by definition of $\phi_{e_H}$ |

Thus the morphism property is demonstrated, and $\phi_{e_H}$ is seen to be a homomorphism.

$\blacksquare$

## Sources

- 1967: George McCarty:
*Topology: An Introduction with Application to Topological Groups*... (previous) ... (next): Chapter $\text{II}$: Groups: Morphisms