Right and Left Regular Representations in Topological Group are Homeomorphisms

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

Let $g \in G$ be an element of $G$.

Then the left and right regular representations with respect to $g$:
 * $L_g: \left({G, \tau}\right) \to \left({G, \tau}\right)$

and:
 * $R_g: \left({G, \tau}\right) \to \left({G, \tau}\right)$

are homeomorphisms.

Proof
Let:
 * $m: \left({G, \tau}\right) \times \left({G, \tau}\right) \to \left({G, \tau}\right)$

be the mapping defined as:
 * $\forall \left({x, y}\right) \in G \times G: m \left({x, y}\right) = x \cdot y$

From the definition of topological group, $m$ is a continuous mapping.

Let $\phi_1: \left({G, \tau}\right) \to \left({G, \tau}\right) \times \left({G, \tau}\right)$ be the mapping defined as:
 * $\phi_1 \left({x}\right) = \left({g, x}\right)$

Let $\phi_2: \left({G, \tau}\right) \to \left({G, \tau}\right) \times \left({G, \tau}\right)$ be the mapping defined as:
 * $\phi_2 \left({x}\right) = \left({x, g}\right)$

Let $\operatorname{pr}_1: \left({G, \tau}\right) \times \left({G, \tau}\right) \to \left({G, \tau}\right)$ be the first projection on $\left({G, \tau}\right) \times \left({G, \tau}\right)$:
 * $\forall \left({x, g}\right) \in G \times G: \operatorname{pr}_1 \left({x, g}\right) = x$

Let $\operatorname{pr}_2: \left({G, \tau}\right) \times \left({G, \tau}\right) \to \left({G, \tau}\right)$ be the second projection on $\left({G, \tau}\right) \times \left({G, \tau}\right)$:
 * $\forall \left({g, x}\right) \in G \times G: \operatorname{pr}_2 \left({g, x}\right) = x$

By inspection, it is seen that:
 * $\phi_1$ is the injective restriction to $g \times G$ of the inverse of $\operatorname{pr}_2$.


 * $\phi_2$ is the injective restriction to $G \times g$ of the inverse of $\operatorname{pr}_1$.

Let $\phi_1 \left({x}\right) = \phi_1 \left({y}\right)$ for some $x, y \in G$.

Then by definition of $\phi_1$:
 * $\left({g, x}\right) = \left({g, y}\right)$

By Equality of Ordered Pairs:
 * $x = y$

and so $\phi_1$ is injective.

Similarly iet $\phi_2 \left({x}\right) = \phi_2 \left({y}\right)$ for some $x, y \in G$.

Then by definition of $\phi_2$:
 * $\left({x, g}\right) = \left({y, g}\right)$

By Equality of Ordered Pairs:
 * $x = y$

and so $\phi_2$ is injective.

Let $\left({g, y}\right) \in G \times G$.

Then:
 * $\exists y \in G: \phi_1 \left({y}\right) = \left({g, y}\right)$

and so $\phi_1$ is surjective.

Similarly let $\left({y, g}\right) \in G \times G$.

Then:
 * $\exists y \in G: \phi_2 \left({y}\right) = \left({y, g}\right)$

and so $\phi_2$ is surjective.

Thus as $\phi_1$ and $\phi_2$ are both injective and surjective, they are by definition both bijections.

From Projection from Product Topology is Open, both $\operatorname{pr}_1$ and $\operatorname{pr}_2$ are open.

From Bijection is Open iff Inverse is Continuous:


 * $\phi_1$ and $\phi_2$ are continuous.

From Composite of Continuous Mappings is Continuous:
 * $m \circ \phi_1: \left({G, \tau}\right) \to \left({G, \tau}\right)$

and:
 * $m \circ \phi_2: \left({G, \tau}\right) \to \left({G, \tau}\right)$

are continuous mappings.

By definition of left regular representation:
 * $\forall x \in G: L_g \left({x}\right) = g \cdot x = m \left({g, x}\right) = m \circ \phi_1 \left({x}\right)$

Similarly, by definition of right regular representation:
 * $\forall x \in G: R_g \left({x}\right) = x \cdot g = m \left({x, g}\right) = m \circ \phi_2 \left({x}\right)$

Thus $L_g$ and $R_g$ are continuous.

By definition of inverse element of the group $G$:
 * $L_{g^{-1} } \left({x}\right) = g^{-1} \cdot x$
 * $R_{g^{-1} } \left({x}\right) = x \cdot g^{-1}$

Thus for all $x \in G$:
 * $L_g \circ L_{g^{-1} } \left({x}\right) = g \cdot \left({g^{-1} \cdot x}\right) = x$
 * $L_{g^{-1} } \circ L_g \left({x}\right) = g^{-1} \cdot \left({g \cdot x}\right) = x$

and similarly:
 * $R_g \circ R_{g^{-1} } \left({x}\right) = \left({x \cdot g^{-1} }\right) \cdot g = x$
 * $R_{g^{-1} } \circ R_g \left({x}\right) = \left({x \cdot g}\right) \cdot g^{-1} = x$

So:
 * $L_g \circ L_{g^{-1} } = L_{g^{-1} } \circ L_g = I_G$
 * $R_g \circ R_{g^{-1} } = R_{g^{-1} } \circ R_g = I_G$

where $I_G$ is the identity mapping on $G$.

Thus by definition, $L_{g^{-1} }$ and $R_{g^{-1} }$ are the inverses of $L_g$ and $R_g$ respectively.

As $L_{g^{-1} }$ and $R_{g^{-1} }$ are themselves regular representations with respect to elements of $G$, they are themselves continuous bijections.

So $L_g$ and $R_g$ are continuous bijections whose inverses are likewise continuous bijections.

Hence, by definition, $L_g$ and $R_g$ are homeomorphisms.