For a metric $d$ on a group $G$, why do $d$ and $d^{-1}$ generate the same topology.

Question

Let $d$ be a metric on a group $G$ and define $d^{-1}$ by $d^{-1}(x,y)=d(x^{-1},y^{-1})$. Why do $d$ and $d^{-1}$ generate the same metric topology on $G$?

Let $g \in G$ and $\epsilon >0$. Let $B_d=B_d(g,\epsilon)=\{h \in G : d(g,h) < \epsilon\}$. I need to find a $d^{-1}$ ball inside of $B_d$ but am having trouble. Did I set the problem up correctly?

Any hints/solutions are appreciated here.

Answer 1

I believe you are missing an assumption. Your metric shouldn't be any old metric on the set of elements of $G$, it needs to respect the group structure.

For example, taking the inverse of an element should be continuous function with regards to $d$. How could you use that continuity to find a $\epsilon_2$ such that $B_{d^{-1}}(g,\epsilon_2)\subset B_d(g,\epsilon)$?