In motivating $A\_\infty$-spaces to my students I'm going to insist on the homotopy invariance of the notion, saying that "being $A\_\infty$ is the homotopy invariant version of being a topological monoid" and to stress this I'd like to say that if $X$ is a topological monoid and $Y$ is a space homotopy equivalent to $X$ then $Y$ will carry an $A\_\infty$-structure making it equivalent to $X$ as an $A\_\infty$-space, but in general not a topological monoid structure with this property. But at this point I see to my shame that I miss an explicit example of this! 


Clearly the most dramatic example would be that of a space $Y$ which is homotopy equivalent to a topological monoid $X$, but such $Y$ carries no topological monoid structure at all, not to have to go into the equivalence issue. For a while I thought the closed interval could be an example of this (double shame: there are at least two very simple and well known topological monoid structures on $[0,1]$!), so I'm completely without examples, and I do not either know if such a space $Y$ does actually exist at all.


Any suggestion?


edit: despite I originally formulated my question in the most dramatic possible form, an example where, given a homotopy equivalence $f:Y\to X$ there is no monoid structure on $Y$ such that $\pi\_0(f)$ is an isomorphism of monoids $\pi\_0(Y)\to \pi\_0(X)$ is even better for what I need to explain, namely that going from monoids to $A\_\infty$-spaces not only $Y$ is naturally endowed with an $A\_\infty$-space structure, but $f$ is promoted to an equivalence of $A\_\infty$-spaces. So I will now leave the original question open as a general topology question which may have its interest in its own (despite it is admittedly an odd question), while for myself I'll be perfectly satisfied with the very nice answer by Tyler below.



Spaces with no topological monoid structure which are homotopy equivalent to topological monoids
We can modify Neil's argument in the other thread to give an example of a contractible space with no monoid structure.


Let T be a tree with the following property: 


For each point x in T, there are at least two components of T \ x which contain an at least trivalent vertex.


In particular, the complement of x must have at least two components, so T cannot contain any 1-valent vertices.


For example, let T be the universal cover of the theta graph.


Now suppose that T has a monoid structure. Using Neil's argument, we can get some invertibility results. Namely, let x and y be points in two different components of the complement of the identity. Then there is a path from (e,y) to (x,y) to (x,e) in $T\times T$ which holds y fixed in the first half and x in the second half. The image of this path must pass through e, so either x or y is invertible. This implies that every point in all but one of the components of the complement of e is invertible. Then there is an at least trivalent vertex g distinct from e which is invertible and a non-constant path from $g^{-1}$ to e through invertible elements. This gives a homotopy of homeomorphisms from T to T by left multiplication. The image of g at time 0 is e; at time 1 it is g. This means at some time between them, g must be taken to an internal point of an edge of T; no homeomorphism can do that.


Such a T is homotopy equivalent to a point but can have no monoid structure.


