Let $(X, x\_0)$ be a based topological space, and $\Omega X$ its based loop space. The group of path components of $\Omega X$ is $\pi\_0(\Omega X) = \pi\_1(X, x\_0)$. For brevity, let's call this group $G$. Since $\Omega X$ is a loop space, all of its components are homotopy equivalent, and so if we define $\Omega\_1 X$ to be the component associated to $1 \in G$ (i.e., the null-homotopic loops), then there is a homotopy equivalence 


$$\varphi: G \times \Omega\_1 X \to \Omega X.$$


Explicitly, $\varphi(g, f) = r(g)\*f$, where $r: G \to \Omega X$ is a section of $\pi\_0$, and $\*$ is concatenation of loops. My question is about the extent to which this map is a loop map (or $A\_\infty$ map).


We can equip the domain of $\varphi$ with the structure of an $H$-space with multiplication


$$(g\_1, f\_1) \cdot (g\_2, f\_2) := (g\_1 g\_2, r(g\_2)^{-1}\*f\_1 \* r(g\_2) \* f\_2);$$


i.e., a sort of semidirect product structure where $G$ acts on $\Omega\_1 X$ by conjugation of loops (via $r$). Then $\varphi$ is an $H$-map with respect to this multiplication: i.e., it is multiplicative up to homotopy. However, the domain is not a strictly associative monoid with respect to the multiplication $\cdot$, nor is $\varphi$ a strict morphism.


So, my question is: is there an $A\_\infty$ structure on $G \times \Omega\_1 X$ agreeing the multiplicative structure described above (up to homotopy) and a map $\Phi: G \times \Omega\_1 X \to \Omega X$, homotopic $\varphi$, which is an $A\_\infty$ map?



Components of a loop space, semidirect products, and multiplicativity
I think the answer is ' yes' for trivial reasons. If the loop space were a topological group the map $\varphi$ would be a homeomorphisms and the group structure on the target could be transferred to the source. Now, any space has models for the loop space which are honest topological groups, so you can transfer the result along these homotopy equivalences.


