Epimorphism from Real Numbers to Circle Group

Theorem
Let $\struct {K, \times}$ be the circle group, that is:
 * $K = \set {z \in \C: \cmod z = 1}$

under complex multiplication.

Let $f: \R \to K$ be the mapping from the real numbers to $K$ defined as:
 * $\forall x \in \R: \map f x = \cos x + i \sin x$

Then $f: \struct {\R, +} \to \struct {K, \times}$ is a group epimorphism.

Its kernel is:
 * $\map \ker f = \set {2 \pi n: n \in \Z}$

Proof
$f$ is a surjection from ...

Then:

So $f$ is a (group) homomorphism.

Thus $f$ is seen to be a surjective homomorphism.

Hence, by definition, it is a (group) epimorphism.

From Cosine of Multiple of Pi:
 * $\forall n \in \Z: \cos n \pi = \paren {-1}^n$

and from Sine of Multiple of Pi:
 * $\forall n \in \Z: \sin n \pi = 0$

From Sine and Cosine are Periodic on Reals, it follows that these are the only values of $\Z$ for which this holds.

For $\cos x + i \sin x = 1 + 0 i$ it is necessary that:
 * $\cos x = 1$
 * $\sin x = 0$

and it can be seen that the only values of $x$ for this to happen is:
 * $x \in \set {2 \pi n: n \in \Z}$

Hence, by definition of kernel:
 * $\map \ker f = \set {2 \pi n: n \in \Z}$