Left shift continuity

Question

Let $G$ be a topological group. It says that left shift $L_g: G \rightarrow G$ ($ x\mapsto g\cdot x$) is continious map, but i am confused about proving it. As I expect, it somehow follows from the continuity of multiplication. Can someone help me? Many-many thanks!

Answer 1

Let $h\in G$ and $U\subseteq G$ open. We want to prove that $L_h^{-1}(U)$ is open.

We know $\mu:G\times G\to G$ given by $\mu(g_1,g_2)=g_1g_2$ is continuous, so $\mu^{-1}(U)$ is open in $G\times G$. By definition of product topology, there are $\{U_\alpha\}_{\alpha\in I}$ and $\{V_\alpha\}_{\alpha\in I}$ open in $G$ such that: $\mu^{-1}(U)=\bigcup_{\alpha \in I}U_\alpha\times V_\alpha$

Let $g\in L_h^{-1}(U)$. Since $hg\in U$ we have $(h,g)\in \mu^{-1}(U)$, hence $(h,g)\in U_\alpha\times V_\alpha$ for some $\alpha \in I$. Then $\{h\}\times V_\alpha\subseteq U_\alpha\times V_\alpha \subseteq \mu^{-1}(U)$. Therefore $hk\in U$ for all $k\in V_\alpha$, and $g\in V_\alpha\subseteq L_h^{-1}(U)$. Since $V_\alpha$ is open, $L_h^{-1}(U)$ is open.

Answer 2

The function $\mu: G \times G \rightarrow G, (g,x) \mapsto gx$ is continuous, where $G \times G$ is continuous in the product topology. Fix $g \in G$. Then the restriction of $\mu$ to $\{g\} \times G \rightarrow G$ is still continuous.

Also, the map $x \mapsto (g,x)$ is homeomorphism from $G$ onto $\{g\} \times G$. This is clear, because the open sets in $\{g\} \times G$ are all of the form $\{g\} \times E$, where $E$ is open in $G$, so the bijection $x \mapsto (g,x)$ induces a bijection between the open sets of $G$ and open sets of $\{g\} \times G$.

So, the function $G \rightarrow G$ given by $x \mapsto gx$ is continuous, because it is the composition of the continuous functions $G \rightarrow \{g\} \times G \rightarrow G$.