Exercise TG.4

Proof. Clearly $G$ is meant to be a topological group. Let $e$ be the identity element of $G$. Define the function $f_{\alpha }^{\prime }:G\to G$ by $f_{\alpha }^{\prime }(x)={\alpha }^{-1}\cdot x$. For any any $x\in G$ we then have

and

This shows that both $f_{\alpha }^{\prime }\circ f_{\alpha }=i_G$ and $f_{\alpha }\circ f_{\alpha }^{\prime }=i_G$ so that $f_{\alpha }^{\prime }$ is both a left inverse and a right inverse for $f_{\alpha }$ (see Exercise 2.5). Hence $f_{\alpha }$ is bijective and ${f_{\alpha }}^{-1}=f_{\alpha }^{\prime }$ by Exercise 2.5 part (e). An analogous argument shows $g_{\alpha }$ is bijective and that $g_{\alpha }^{\prime }(x)=x\cdot {\alpha }^{-1}$ is its inverse function.

To show that they are homeomorphisms, we note that the group operation is a continuous function since $G$ is a topological group. We then have that the operation is continuous in each variable separately by Exercise 18.11. From this it clearly follows that both $f_{\alpha }$ and $g_{\alpha }$ are continuous since $f_{\alpha }(x)=\cdot (\alpha \times x)$ and $g_{\alpha }(x)=\cdot (x\times \alpha )$. Similarly ${f_{\alpha }}^{-1}$ and ${g_{\alpha }}^{-1}$ are continuous since we have ${f_{\alpha }}^{-1}(x)=f_{\alpha }^{\prime }(x)=\cdot ({\alpha }^{-1}\times x)$ and ${g_{\alpha }}^{-1}(x)=g_{\alpha }^{\prime }(x)=\cdot (x\times {\alpha }^{-1})$. This shows that both $f_{\alpha }$ and $g_{\alpha }$ are homeomorphisms by definition.

Now, to show that $G$ is a homogeneous space, consider any points $x_0,y_0\in G$. Set $\alpha =y_0\cdot {x_0}^{-1}$, noting that of course $\alpha \in G$. We then claim that $f_{\alpha }$ as defined above is a homeomorphism that carries $x_0$ to $y_0$. Of course we already showed above that $f_{\alpha }$ is a homeomorphism, so all that remains is to show that $f_{\alpha }(x_0)=y_0$. To this end we have

which shows the desired result. □