Equivalence of Definitions of Topological Group

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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:

\((1)\)   $:$   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:

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

Then $\phi'$ 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.

$\blacksquare$