Constant Mapping to Identity is Homomorphism

Group Homomorphism
Let $\left({G_1, \circ_1}\right)$ and $\left({G_2, \circ_2}\right)$ be groups with identities $e_1$ and $e_2$ respectively.

Let $\phi_e: \left({G_1, \circ_1}\right) \to \left({G_2, \circ_2}\right)$ be the constant mapping defined as:
 * $\forall x \in G_1: \phi_e \left({x}\right) = e_2$

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

Ring Homomorphism
Let $\left({R_1, +_1, \circ_1}\right)$ and $\left({R_2, +_2, \circ_2}\right)$ be rings with zeroes $0_1$ and $0_2$ respectively.

Let $\theta_0: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be the constant mapping defined as:
 * $\forall x \in R_1: \theta_0 \left({x}\right) = 0_2$

Then $\theta_0$ is a ring homomorphism whose image is $\left\{{0_2}\right\}$ and whose kernel is $R_1$.

Proof for Group Homomorphism
Let $x, y \in G_1$.

Then:

So $\phi_e$ is a group homomorphism.

The results about image and kernel follow directly by definition.

Proof for Ring Homomorphism
The additive groups of $\left({R_1, +_1, \circ_1}\right)$ and $\left({R_2, +_2, \circ_2}\right)$ are $\left({R_1, +_1}\right)$ and $\left({R_2, +_2}\right)$ respectively.

Their identities are $0_1$ and $0_2$ respectively.

Thus from the proof for Group Homomorphism we have that $\theta_0: \left({R_1, +_1}\right) \to \left({R_2, +_2}\right)$ is a group homomorphism:
 * $\theta_0 \left({x +_1 y}\right) = \theta_0 \left({x}\right) +_2 \theta_0 \left({y}\right)$

Then we have:

The results about image and kernel follow directly by definition.