If $(X,e)$ is a based topological space, then the James construction on $(X,e)$ is the free topological monoid with unit $e$: $J(X)=\coprod_{n\geq 1}X^n /\sim $ where $(x_1,...x_{j-1},e,x_j,...,x_n)\sim (x_1,...x_{j-1},x_j,...,x_n)$. James famously proved that if $X$ is a countable cell-complex, then $J(X)$ is homotopy equivalent to $\Omega\Sigma X$.

Milnor gives a similar construction in the unpublished paper "The construction $FK$" where $FK=\bigcup_{n}F(K_n)$ is the free simplicial group generated by a semi-simplicial complex $K$. Working in the category of simplicial complexes, Milnor shows that $FK$ is a loop space for the suspension of $X$.

I would like to know if the topological group version of these results still holds: let $X$ be a countable connected CW-complex and let $X^{-1}$ be a homeomorphic copy of $X$. There is a natural map $R:J(X\vee X^{-1})\to F(X)$ to the free group on the set $X$. If we view the free group $F(X)$ on the underlying set of $X$ as the quotient space of $J(X\vee X^{-1})$, then $F(X)$ becomes the free Graev topological group $F_G(X)$ on $X$. Since the free topological group functor can be tricky to work with, it is not clear to me if $F_G(|K|)\simeq |FK|$ when $K$ is a countable semi-simplicial complex.

Note: countability of $X$ is necessary to ensure that this construction of $F_G(X)$ actually yields the free topological group.

Question: Is $F_G(X)$ homotopy equivalent to $\Omega \Sigma X$?