Mapping to Identity is Unique Constant Homomorphism

Theorem
Let $\left({G, \circ}\right)$ and $\left({H, *}\right)$ 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:
 * $\phi_{e_H}: G \to H: \forall g \in G: \phi_{e_H} \left({g}\right) = e_H$

Proof
Let $h \in H$ such that $\phi_h: G \to H$ is a (group) homomorphism, where $\phi_h$ is defined as:
 * $\forall g \in G: \phi_h \left({g}\right) = h$

Then from Group Homomorphism Preserves Identity:
 * $\phi_h \left({e_G}\right) = e_H$

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 $x, y \in G$.

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