Imaginary Part as Mapping is Endomorphism for Complex Addition

Theorem
Let $\left({\C, +}\right)$ be the additive group of complex numbers.

Let $\left({\R, +}\right)$ be the additive group of real numbers.

Let $f: \C \to \R$ be the mapping from the real numbers to the complex numbers defined as:
 * $\forall z \in \C: f \left({z}\right) = \Im \left({z}\right)$

where $\Im \left({z}\right)$ denotes the imaginary part of $z$.

Then $f: \left({\C, +}\right) \to \left({\R, +}\right)$ is a group epimorphism.

Its kernel is the set:
 * $\ker \left({f}\right) = \R$

of (wholly) real numbers.

Proof
For all $x \in \R$, there exists a complex number $x + i y$ where $y$ is another real number.

So $f$ is a surjection.

Let $z_1, z_2 \in \C$.

Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$.

Then:

So $f$ is a group homomorphism.

Thus $f$ is a surjective group homomorphism and therefore by definition a group epimorphism.

Finally:
 * $\forall y \in \R: \Im \left({x + 0 i}\right) = 0 i = 0$

It follows from Complex Addition Identity is Zero that:
 * $\ker \left({f}\right) = \left\{{x: x \in \R}\right\} = \R$