# The group analogue of the James construction

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Milnor gives a similar construction in the unpublished paper "The construction $FK$" where $FK=\bigcup_{n}F(K_n)$ is the free simplicial group generated by a semi-simplicial complex $K$.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

• $F$ is defined over simplicial sets whereas $J$ is defined over topologial spaces. It is classic that if $X$ is connected then $|F(X)|$ is homotopy equivalent to $J(|X|)$ which is a model for $\Omega\Sigma |X|$. – user51223 Jun 29 '17 at 16:24
• @user51223 I'll clarify my notation. I use $F_G(X)$ to define the free Graev topological group on a topological space $(X,e)$. My questions is if $F_G(X)\simeq \Omega\Sigma X$ for a space $X$. – Jeremy Brazas Jun 29 '17 at 16:43
• Can you please state the definition of $F_G$ more clearly? Do you write $X^{-1}$ to mean that the elements of this space are formal inverses to those of $X$ in the free monoid on $X$, so that $F(X\vee X^{-1})$ is a group?! – user51223 Jun 30 '17 at 12:42
• To get a free group on $X$ (where $e$ is the identity), you first need formal inverses in the monoid, hence the use of $J(X\vee X^{-1},e)$. Elements here can be thought of as unreduced words. Collapsing according to word reduction yields the free group. – Jeremy Brazas Jun 30 '17 at 15:27