Mapping to Identity is Unique Constant Homomorphism

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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:

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


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.