Let $G$ be a topological group, and let $F$ be the Markov free topological group over $G$. We define an action of $G$ on $F$ as follows: $G\times F\rightarrow F$, $^{g}(g\_{1}^{\epsilon\_{1}}...g\_{n}^{\epsilon\_{n}})$=$(^{g}g\_{1})^{\epsilon\_{1}}...(^{g}g\_{n})^{\epsilon\_{n}}$.
**Question:** Is the above-defined action, a contiuous action? Why? (Note that, here $G$ acts on itself by conjugation.)



a free topological group as a topological module
I don't know the answer to the general question but the answer seems to be "yes" when $G$ is compact. Here is a straightforward argument.


The corresponding action of $G$ on the free topological monoid $M=\coprod\_{n\geq 0}(G\sqcup G^{-1})^n$ is certainly continuous. When $G$ is compact $F(G)$ is the topological quotient of $M$ via word reduction $r:M\to F(G)$. When the action of $G$ on itself is conjugation, we have $^{g}(r(g\_{1}^{\epsilon\_1}...g\_{n}^{\epsilon\_n}))=r(^{g}(g\_{1}^{\epsilon\_1}...g\_{n}^{\epsilon\_n}))$ for any $g\_{1}^{\epsilon\_1}...g\_{n}^{\epsilon\_n}\in M$. The action is continuous since $id\times r:G\times M\to G\times F(G)$ is a topological quotient map (compactness of $G$ helps out here too).


This argument should also work when $G$ is a $k\_{\omega}$-space (inductive limit of nested compact subspaces) but doesn't extend to the general case since $r:M\to F(G)$ is not quotient when $G=\mathbb{Q}$ is the rationals.


