Equivalence of Definitions of Topological Group

Theorem

Let $\struct {G, \odot}$ be a group.

On its underlying set $G$, let $\struct {G, \tau}$ be a topological space.

The following definitions of the concept of Topological Group are equivalent:

Definition 1

$\struct {G, \odot, \tau}$ is a topological group if and only if the following conditions are fulfilled:

\((1)\)	$:$		Continuous Group Product					$\odot: \struct {G, \tau} \times \struct {G, \tau} \to \struct {G, \tau}$ is a continuous mapping
\((2)\)	$:$		Continuous Inversion Mapping					$\iota: \struct {G, \tau} \to \struct {G, \tau}$ such that $\forall x \in G: \map \iota x = x^{-1}$ is also a continuous mapping

where $\struct {G, \tau} \times \struct {G, \tau}$ is considered as $G \times G$ with the product topology.

Definition 2

Let the mapping $\psi: \struct {G, \tau} \times \struct {G, \tau} \to \struct {G, \tau}$ be defined as:

$\map \psi {x, y} = x \odot y^{-1}$

$\struct {G, \odot, \tau}$ is a topological group if and only if:

$\psi$ is a continuous mapping

where $\struct {G, \tau} \times \struct {G, \tau}$ is considered as $G \times G$ with the product topology.

Proof

Definition 1 implies Definition 2

Let $\struct {G, \odot, \tau}$ be a topological group by Definition 1.

Let $\phi: \struct {G, \tau} \to \struct {G, \tau}$ be the mapping defined as:

$\forall x \in G: \map \phi x = x^{-1}$

By definition:

$\odot: \struct {G, \tau} \times \struct {G, \tau} \to \struct {G, \tau}$ is a continuous mapping

$\phi: \struct {G, \tau} \to \struct {G, \tau}$ is a continuous mapping

Let $\phi': G \times G \to G \times G$ be defined by:

$\map {\phi'} {x, y} = \tuple {x, \map \phi y}$

Let $\phi_1: G \times G \to G$ be defined as:

$\map {\phi_1} {x, y} = x$

Let $\phi_2: G \times G \to G$ be defined as:

$\map {\phi_2} {x, y} = y^{-1}$

Then for arbitrary open set $V \subset G$, both sets

$\phi_1^{-1} \sqbrk V = V \times G$

and

$\phi_2^{-1} \sqbrk V = G \times \phi^{-1} \sqbrk V$

are open in $G \times G$, thus $\phi_1$ and $\phi_2$ are continuous.

Because:

$\map {\phi'} {x, y} = \tuple {\map {\phi_1} x, \map {\phi_2} y}$

by Continuous Mapping to Product Space, $\phi'$ is continuous.

Let the mapping $\psi: \struct {G, \tau} \times \struct {G, \tau} \to \struct {G, \tau}$ be defined as:

$\psi = \odot \circ \phi'$

where $\circ$ denotes composition of mappings.

By Composite of Continuous Mappings is Continuous, $\psi$ is continuous.

Then:

\(\ds \map \psi {x, y}\)

\(=\)

\(\ds \map {\paren {\odot \circ \phi'} } {x, y}\)

Definition of $\psi$

\(\ds \)

\(=\)

\(\ds \map \odot {\map {\phi'} {x, y} }\)

Definition of Composition of Mappings

\(\ds \)

\(=\)

\(\ds \map \odot {x, \map \phi y}\)

Definition of $\phi'$

\(\ds \)

\(=\)

\(\ds \map \odot {x, y^{-1} }\)

Definition of $\phi$

\(\ds \)

\(=\)

\(\ds x \odot y^{-1}\)

Definition of $\odot$

demonstrating that:

$\map \psi {x, y} = x \odot y^{-1}$

Thus $\struct {G, \odot, \tau}$ is a topological group by Definition 2.

$\Box$

Definition 2 implies Definition 1

Let $\struct {G, \odot, \tau}$ be a topological group by Definition 2.

Let $e$ be the identity of $G$.

Let the mapping $\psi: \struct {G, \tau} \times \struct {G, \tau} \to \struct {G, \tau}$ be defined as:

$\forall \tuple {x, y} \in G \times G: \map \psi {x, y} = x \odot y^{-1}$

By definition of $\struct {G, \odot, \tau}$, $\psi$ is continuous.

By Continuous Mapping to Product Space $\psi$ is continuous in each variable.

Let $\phi: \struct {G, \tau} \to \struct {G, \tau}$ be the mapping defined as:

$\forall x \in G: \map \phi x = x^{-1}$

Since $\map \phi x = \map \psi {e, x}$, it follows that $\phi$ is continuous.

Let $\phi': G \times G \to G \times G$ be defined by:

$\forall \tuple {x, y} \in G \times G: \map {\phi'} {x, y} = \tuple {x, \map \phi y}$

By Continuous Mapping to Product Space, $\phi'$ is continuous.

$\odot$ is the composition of $\psi$ with $\phi'$.

Thus by Composite of Continuous Mappings is Continuous, $\odot$ is continuous.

Thus $\struct {G, \odot, \tau}$ is a topological group by Definition 1.

$\blacksquare$