Quotient Theorem for Group Homomorphisms/Examples/Real to Complex Numbers under e^2 pi i x

Example of Use of Quotient Theorem for Group Homomorphisms
Let $\struct {\R, +}$ denote the additive group of real numbers.

Let $\struct {\C_{\ne 0}, \times}$ denote the multiplicative group of complex numbers.

Let $\phi: \struct {\R, +} \to \struct {\C_{\ne 0}, \times}$ be the homomorphism defined as:
 * $\forall x \in \R: \map \phi x = e^{2 \pi i x}$

Then $\phi$ can be decomposed into the form:
 * $\phi = \alpha \beta \gamma$

in the following way:


 * $\alpha: \struct {K, \times} \to \struct {\C_{\ne 0}, \times}$ is defined as:
 * $\forall z \in K: \map \alpha z = z$


 * where $\struct {K, \times}$ denotes the circle group:
 * $K = \set {z \in \C: \cmod z = 1}$
 * $\times$ is the operation of complex multiplication


 * $\beta: \hointr 0 1 \to K$ is defined as:
 * $\forall x \in \hointr 0 1: \map \beta x = e^{2 \pi i x}$
 * where $\hointr 0 1$ denotes the right half-open real interval $\set {x \in \R: 0 \le x < 1}$


 * $\gamma: \R \to \hointr 0 1$ is defined as:
 * $\forall x \in \R: \map \gamma x = \fractpart x$
 * where $\fractpart x$ is the fractional part of $x$:
 * $\fractpart x := x - \floor x$

Proof
It is first demonstrated that $\phi$ is a homomorphism:

We have that:

By Group Homomorphism Preserves Identity it is confirmed that $1$ is the identity of $\struct {\C_{\ne 0}, \times}$.

Now we can establish what the kernel of $\phi$ is:

Thus:
 * $\map \ker \phi = \Z$

where $\Z$ denotes the set of integers.

Next we establish what the image of $\phi$ is:

Thus, from the Quotient Theorem for Group Homomorphisms, $\phi$ can be decomposed into:
 * $\phi = \alpha \beta \gamma$

where:
 * $\alpha: K \to \C_{\ne 0}$, which is a monomorphism
 * $\beta: \R / \Z \to K$, which is an isomorphism
 * $\gamma: \R \to \R / \Z$, which is an epimorphism.

As $\beta$ is an isomorphism, $\beta$ is also a bijection and so $\R / \Z = \Preimg \beta$ can be deduced:

We have specifically selected $\hointr 0 {2 \pi}$ as the image of the principal argument of $\Ln z$.

Other half-open real intervals whose length is $2 \pi$ work equally well, for example $\hointl {-\pi} \pi$.

Thus:

Thus we have:
 * $\Preimg \beta = \R / \Z = \hointr 0 1$

and the result follows.