TITLE: Is an A-infinity thing the same the same as strict thing viewed through a homotopy equivalence?
QUESTION [16 upvotes]: If I have a topological monoid ($X,\mu$), a space $Y$ and a homotopy equivalence $f,g$ between them, then $Y$ has the structure of an $A_\infty$ space defined 'pointwise' by
$$ y_1 * y_2 := g \left(\mu(f(y_1),f(y_2))\right) $$
Also, I have heard someone say the reverse as "Every $A_\infty$-space is (weakly?) homotopy equivalent to a topological monoid."
My question is this: In how far is this the general case? Can I define/think of $A_\infty$ structures as strict structures as viewed through the distorting glasses of a homotopy equivalence? When is an $A_\infty$ structure of this type - i.e. is there always an equivalent strict version? (I guess no in general. Can you tell me more?) 
FYI: I started out with the question: Is the based loop space of a space $X$ always on the other side of a homotopy equivalence of a (strict) topological group?
thanks,
a.

REPLY [6 votes]: Some ancient history: The first proof that the loop space of any reasonable space is equivalent to a topological group is due to John Milnor, in his 1956 Annals of Math. paper  ``Construction of universal bundles, I'' (I'm sure its hypotheses are removable nowadays.)
Some slightly less ancient history is Theorem 13.5 in my 1972 Springer Volume ``The Geometry of Iterated Loop spaces'', http://www.math.uchicago.edu/~may/BOOKS/geom_iter.pdf on my web page. An $A_{\infty}$-space is a $\mathcal{C}$-space $X$ for any $A_{\infty}$ operad $\mathcal{C}$. In the notation there, I construct a topological monoid $B(M,C,X)$, a $\mathcal{C}$-space $B(C,C,X)$ and a pair of $\mathcal{C}$-maps $B(C,C,X) \longrightarrow X$ and $B(C,C,Z)\longrightarrow B(M,C,X)$.  The first is always a homotopy
equivalence and the second is always a weak homotopy equivalence, although I only proved that there when $X$ is connected.  This is the standard construction that Tyler described in more detail. (I don't think I was yet on Overflow when the question was first asked.)