Equivalence of Definitions of Topological Group

Theorem
Let $\left({G, \odot}\right)$ be a group.

On its underlying set $G$, let $\left({G, \tau}\right)$ be a topological space.

Definition 1 implies Definition 2
Let $\left({G, \odot, \tau}\right)$ be a topological group by Definition 1.

Let $\phi: \left({G, \tau}\right) \to \left({G, \tau}\right)$ be the mapping defined as:
 * $\forall x \in G: \phi \left({x}\right) = x^{-1}$

By definition:
 * $\odot: \left({G, \tau}\right) \times \left({G, \tau}\right) \to \left({G, \tau}\right)$ is a continuous mapping
 * $\phi: \left({G, \tau}\right) \to \left({G, \tau}\right)$ is a continuous mapping

Let $\phi': G \times G \to G \times G$ be defined by
 * $\phi' \left({x, y}\right) = \left({x, \phi \left({y}\right) }\right)$.

Let $\phi_1: G \times G \to G$ be defined as
 * $\phi_1 \left({x, y}\right) = x$.

Let $\phi_2: G \times G \to G$ be defined as
 * $\phi_2 \left({x, y}\right) = y^{-1}$.

Then for arbitrary open set $V \subset G$, both sets
 * $\phi_{1}^{-1} \left(V\right) = V \times G$

and
 * $\phi_{2}^{-1} \left(V\right) = G \times \phi^{-1}\left(V\right)$

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

Since
 * $\phi' \left({x,y}\right) = \left( {\phi_1 \left(x)\right), \phi_2 \left(y\right)} \right)$,

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

Let the mapping $\psi: \left({G, \tau}\right) \times \left({G, \tau}\right) \to \left({G, \tau}\right)$ 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:
 * $\psi: \left({x, y}\right) = x \odot y^{-1}$

Thus $\left({G, \odot, \tau}\right)$ is a topological group by Definition 2.

Definition 2 implies Definition 1
Let $\left({G, \odot, \tau}\right)$ be a topological group by Definition 2.

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

Let the mapping $\psi: \left({G, \tau}\right) \times \left({G, \tau}\right) \to \left({G, \tau}\right)$ be defined as:
 * $\forall \left({x, y}\right) \in G \times G: \psi: \left({x, y}\right) = x \odot y^{-1}$

By definition of $\left({G, \odot, \tau}\right)$, $\psi$ is continuous.

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

Let $\phi: \left({G, \tau}\right) \to \left({G, \tau}\right)$ be the mapping defined as:
 * $\forall x \in G: \phi \left({x}\right) = x^{-1}$

Since $\phi \left({x}\right) = \psi \left({e, x}\right)$, it follows that $\phi$ is continuous.

Let $\phi': G \times G \to G \times G$ be defined by:
 * $\forall \left({x, y}\right) \in G \times G: \phi' \left({x, y}\right) = \left({x, \phi \left({y}\right) }\right)$

Then $\phi'$ is continuous.

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

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

Thus $\left({G, \odot, \tau}\right)$ is a topological group by Definition 1.