Let $G=\text{Gal}(\overline{\mathbf Q}/\mathbf Q)$, and for each prime $p$, choose an embedding $\overline{\mathbf Q} \hookrightarrow \overline{\mathbf Q_p}$. Let $\sigma_p$ be a choice of Frobenius in $\text{Gal}(\overline{\mathbf Q_p}/{\mathbf Q_p})$ and denote also by $\sigma_p$ its image in $G$ by the restriction map $\text{Gal}(\overline{\mathbf Q_p}/{\mathbf Q_p}) \hookrightarrow G$.
Let $H$ be the subgroup of $G$ generated by the $\sigma_p$'s. By the Chebotarev density theorem, $H$ is dense in $G$.
Question: Is $H$ free on the generators $\{\sigma_p\}$?
It seems likely to me that this is true. Any relation between the $\{\sigma_p\}$'s in $G$ would descend to a relation between the (chosen) Frobenii of any finite Galois extension of $\mathbf Q$. It seems unlikely that there might be such a universal relation.
(Interestingly, there is a relation if we include also the Frobenius at $\infty$, namely complex conjugation which satisfies $\sigma^2=1$.)
I am interested in the resulting morphism of profinite groups $\widehat{H} \to G$, where $\widehat{H}$ is the profinite completion of $H$ considered as a discrete group. If $H$ is indeed free on the $\{\sigma_p\}$, then $\widehat{H}$ is profinite-free on the $\{\sigma_p\}$. Is this homomorphism an isomorphism? It seems to me that it should be a bijective homomorphism of profinite groups, but that it is probably not a homeomorphism (since after all, in constructing $\widehat{H}$ we forgot the topology on $H$ induced from $G$)...
Let me alter things a little by taking $G$ to be the Galois group of the maximal extension unram. outside $S$ (a fixed finite set of primes) and $H$ to be the free group generated by symbols $\sigma_p$ for $p \not\in S$. Then we have a morphism $H \to G$ with dense image (send $\sigma_p$ to Frobenius at $p$).
A profinite group is compact, and the morphism $\widehat{H} \to G$ is continuous, with closed and dense image, hence surjective. If it is furthermore bijective, then it is a homeomorphism (source and target are compact).
However, it is not a bijection by any means; it has enormous numbers of relations. For example, the group $G$ has (by Serre's conjecture) has very constrained representations into $GL_2(\overline{\mathbb F_2})$. (These have to correspond to mod $2$ modular forms of level prime to $S$, by Serre's conj.)
These relations are subtle, which is what makes the reciprocity between automorphic forms and Galois representations deep and difficult to prove, but so powerful.