Proving that $f:G\to G $ such that $f(x)=a\cdot x$ is an homeomorphism

Question

I'm having some trouble with the following exercise:

Let $(G,\cdot)$ be a topological group, and let $a\in G$. Consider the function: $$f:G\to G \\ \ \ \ \ \ \ \ \ \ \ x\mapsto a\cdot x$$ Prove that $f$ is an homeomorphism

I was easily able to prove that $f$ is bijective and that $f$ is continuous, but I'm having some trouble proving that it's also open. How can this be done?

Answer 1

Mark Saving's comment is the best next step. However, I wanted to show the map was open directly using the same reasoning involved in the proof of continuity.

Let $U \subseteq G$ be open. We want to show $f(U) = a \cdot U = \{a \cdot x : x \in U\}$ is open.

Since $(G, \cdot)$ is a topological group, the multiplication map $M: G \times G \to G$ is continuous, so the restriction map $M|_{\{a^{-1}\} \times G}:\{a^{-1}\} \times G \to G$ is also continuous. Consider the preimage of $U$ under this map.

Because $f$ is bijective for $a^{-1} \in G$,

$${M|_{\{a^{-1}\}\times G}}^{-1}(U) = \{a^{-1}\}\times f(U)$$

Then $\{a^{-1}\}\times f(U) \subseteq \{a^{-1}\}\times G$ is open, which implies $f(U)$ is open in $G$.