=== 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: {{begin-eqn}} {{eqn | l = \psi \left({x, y}\right)       | r = \left({\odot \circ \phi'}\right) \left({x, y}\right)       | c = Definition of $\psi$ }} {{eqn | r = \odot \left({\phi' \left({x, y}\right)}\right)       | c = Definition of Composition of Mappings }} {{eqn | r = \odot \left({x, \phi \left({y}\right)}\right)       | c = Definition of $\phi'$ }} {{eqn | r = \odot \left({x, y^{-1} }\right)       | c = Definition of $\phi$ }} {{eqn | r = x \odot y^{-1}       | c = Definition of $\odot$ }} {{end-eqn}} demonstrating that: :$\psi: \left({x, y}\right) = x \odot y^{-1}$ Thus $\left({G, \odot, \tau}\right)$ is a topological group by Definition 2. {{qed|lemma}} === 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. {{explain|$\phi'$ 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. {{qed}} Category:Topological Groups i2rz3w2mxd19tziymyu26fajxkfnxu7