Right and Left Multiplications in Topological Group are Homeomorphism

Theorem

Let $(G,\cdot,\tau)$ be a topological group.

Then $L_g:(G,\tau)\to (G,\tau)$ and $R_g:(G,\tau)\to (G,\tau)$; the left and right regular representations respectively, are homeomorphisms.

Proof

From the definition of topological group $m:(G,\tau)\times (G,\tau)\to (G,\tau)$ where $m(x,g)=x\cdot g$ is a continuous mapping.

The mappings $\phi_1:(G,\tau)\to (G,\tau)\times (G,\tau)$ such that $\phi_1(x)=(g,x)$ and $\phi_2:(G,\tau)\to (G,\tau)\times (G,\tau)$ such that $\phi_2(x)=(x,g)$ are inverses of the projections. From Projection from Product Topology is Open it follows that $\phi_1$ and $\phi_2$ are continuous.

Thus, from Composition of Continuous Mappings is Continuous, $m\circ\phi_1:(G,\tau)\to (G,\tau)$ and $m\circ\phi_2:(G,\tau)\to (G,\tau)$ are continuous mappings.

Hence $L_g$ and $R_g$ are continuous, because $m\circ\phi_1(x)=m(g,x)=g\cdot x=L_g(x)$ and $m\circ\phi_2(x)=m(x,g)=x\cdot g=R_g(x)$.

Since $L_g\circ L_{g^{-1}}=L_{g^{-1}}\circ L_g=Id_G$ and $R_g\circ R_{g^{-1}}=R_{g^{-1}}\circ R_g=Id_G$, from the definition of inverse element of the group; $L_g$ and $R_g$ are bijective and their inverses are of the same type: $L_g^{-1}=L_{g^{-1}}$ and $R_g^{-1}=R_{g^{-1}}$.

Hence $\forall g\in G$, $R_g$ and $L_g$ are continuous and; in particular, $\forall g\in G$, $R_g$, $L_g$, $R_{g^{-1}}$ and $L_{g^{-1}}$ are bijective and continuous. Thus $R_g$ and $L_g$ are homeomorphisms.

$\blacksquare$