Note: This is not homework.
Can someone please verify my proof or offer suggestions for improvement?
Let $\alpha$ be an element of a topological group $G$. Show that the map $g_{\alpha}: G \longrightarrow G$ defined by $$g_{\alpha}(x) = x *\alpha$$ is a homeomorphism of G.
Let $j:G \times G \longrightarrow G$ be defined by $$j(x, y) = x*y$$ Since $G$ is a topological group, $j$ is continuous.
Define a map $m_{\alpha}: G \longrightarrow G$ by $$m_{\alpha}(g) = \alpha$$ Clearly, $m_{\alpha}$ is also continuous.
Let $id$ denote the identity map. Then, $id \times m_{\alpha}$ is continuous in the product topology on $G \times G$. Hence, $$j \circ (id \times m_{\alpha})$$ is continuous. Note that $$j \circ (id \times m_{\alpha})(x, y) = j(x, \alpha) = x * \alpha = g_{\alpha}(x)$$ So, $g_\alpha = j \circ (id \times m_{\alpha})$ is continuous. It remains to show that $g_{\alpha}$ is a bijection. It is easy to see that the inverse of $g_{\alpha}$ is $$g_{\alpha}^{-1} = x*\alpha^{-1}$$ Therefore, $g_\alpha$ is invertible and hence bijective. Therefore, $g_\alpha$ is a homeomorphism.
