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.

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:

demonstrating that:
 * $\map \psi {x, y} = x \odot y^{-1}$

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

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.