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Milnor gives a similar construction in the unpublished paper "The construction $FK$" where $FK=\bigcup_{n}F(K_n)$ is the free simplicial group generated by a semi-simplicial complex $K$.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 $\Omega \Sigma X$?

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Milnor gives a similar construction in the unpublished paper "The construction $FK$" where $FK=\bigcup_{n}F(K_n)$ is the free simplicial group generated by a semi-simplicial complex $K$.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: Just as $J(X,e)$ is the free monoid on $X$ where $e$ becomes the identity, the free (Graev) topological group $F_G(X,e)$ is the free group on the set $X$ where $e$ becomes the group identity element. So algebraically, $F_G(X,e)$ is freely generated by the set $X\backslash e$.

To define the topological structure on $F_G(X,e)$, we should restrict to the case where $X$ is a countable connected CW-complex. Let $X^{-1}$ be a homeomorphic copy of $X$ containing formal inverses and basepoint $e^{-1}$. There is a natural reduction map $R:J(X\vee X^{-1},e)\to F_G(X,e)$ taking a word in $J(X\vee X^{-1},e)$ to its reduction in the free group. Now give $F_G(X,e)$ the quotient topology with respect to $R$. In other words, we may construct $F_G(X,e)$ by taking $F_1(X,e)=X\vee X^{-1}$ and inductively gluing copies of $(X\vee X^{-1})^n$ to $F_{n-1}(X,e)$ according to how words are reduced and setting $F_G(X,e)=\bigcup_{n}F_n(X,e)$. In this way, $F_G(X,e)$ becomes a topological group.

$F_G(X,e)$ is called the free (Grave - or based) topological group because it has the expected universal property: any based map $f:(X,e)\to (G,1)$ to a topological group extends uniquely to a continuous homomorphism $\tilde{f}:F_G(X,e)\to G$. The isomorphism type of $F_G(X,e)$ is characterized by this universal property and is also independent of the choice of basepoint $e$.

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

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 $F(X)$ as the quotient space of $J(X\vee X^{-1})$, then $F(X)$ becomes the free topological group on $X$. Since the free topological group functor can be tricky to work with, it is not clear to me if $F(|K|)\simeq |FK|$ when $K$ is a countable semi-simplicial complex.

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

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

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 $\Omega \Sigma X$?