Spaces with no topological monoid structure which are homotopy equivalent to topological monoids - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T14:47:30Z http://mathoverflow.net/feeds/question/91327 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91327/spaces-with-no-topological-monoid-structure-which-are-homotopy-equivalent-to-topo Spaces with no topological monoid structure which are homotopy equivalent to topological monoids domenico fiorenza 2012-03-15T19:38:05Z 2012-03-21T10:43:28Z <p>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! </p> <p>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.</p> <p>Any suggestion?</p> <p>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.</p> http://mathoverflow.net/questions/91327/spaces-with-no-topological-monoid-structure-which-are-homotopy-equivalent-to-topo/91345#91345 Answer by Tyler Lawson for Spaces with no topological monoid structure which are homotopy equivalent to topological monoids Tyler Lawson 2012-03-16T00:49:20Z 2012-03-16T01:52:24Z <p>Let E be a contractible space, and <code>$X = E \coprod \{0\}$</code>. Then there is a homotopy equivalence from $X$ to $\mathbb{Z}/2$ sending all of $E$ to $1$ and $0$ to $0$. The monoid structure on $\mathbb{Z}/2$ lifts to an <code>$A_\infty$</code> structure on $X$.</p> <p>Suppose we could make this come from a topological monoid structure. By checking the induced monoid structure on <code>$\pi_0$</code>, we find that the unit for the monoid structure would have to be in the component of $0$, and hence would have to be equal to $0$. Then for any elements $e$ and $f$ in $E$, their product $ef$ is in the component of $0$ (and hence is $0$). Thus: any two elements in $E$ are both left and right inverse to each other. By the standard uniqueness trick for left-right inverses, this can only happen in the case where $E$ is a singleton.</p> <hr> <p>EDIT: As Benjamin Steinberg points out, this doesn't really answer the question posed because it assumes that we're fixing a homotopy equivalence to some monoid. I don't, at this point, have a better version, but here's an argument based on something that Tom Goodwillie had posted in the comments.</p> <p>Suppose $X$ is disconnected, and can be written as a disjoint union of nonempty open sets $U$ and $V$ where $U$ admits the structure of a topological monoid. Then we can define a topological monoid structure on $X$, extending that on $U$, by picking a point $* \in V$ and defining $uv = vu = v$ if $u \in U, v \in V$, and $v v' = *$ for all $v,v' \in V$. So if we're looking for an example which is, say, locally path-connected, we might as well look for a path-connected example, because if any path component of $X$ admits a topological monoid structure we can extend it to all of $X$. (I've played a little fast and loose with identifying this as a monoid structure, but I think it is correct.)</p> http://mathoverflow.net/questions/91327/spaces-with-no-topological-monoid-structure-which-are-homotopy-equivalent-to-topo/91353#91353 Answer by Neil Strickland for Spaces with no topological monoid structure which are homotopy equivalent to topological monoids Neil Strickland 2012-03-16T06:47:26Z 2012-03-16T06:47:26Z <p>See here: <a href="http://mathoverflow.net/questions/22585" rel="nofollow">http://mathoverflow.net/questions/22585</a></p> <p>The question asked there was different/more general, but the answers actually address the restricted question asked here.</p> http://mathoverflow.net/questions/91327/spaces-with-no-topological-monoid-structure-which-are-homotopy-equivalent-to-topo/91367#91367 Answer by Benjamin Steinberg for Spaces with no topological monoid structure which are homotopy equivalent to topological monoids Benjamin Steinberg 2012-03-16T12:51:29Z 2012-03-17T13:04:05Z <p>It is known that the topological closure of the curve $y=\sin(1/x)$ has no topological monoid structure but I don't know if it is homotopy equivalent to a monoid. </p> <p><b>Edit.</b> It is shown <a href="http://www.springerlink.com/content/76810444m4178420/" rel="nofollow">here</a> that no monoid structure can be put on this space for which multiplication is separately continuous. </p> http://mathoverflow.net/questions/91327/spaces-with-no-topological-monoid-structure-which-are-homotopy-equivalent-to-topo/91814#91814 Answer by Gabriel C. Drummond-Cole for Spaces with no topological monoid structure which are homotopy equivalent to topological monoids Gabriel C. Drummond-Cole 2012-03-21T10:43:28Z 2012-03-21T10:43:28Z <p>We can modify Neil's argument in the other thread to give an example of a contractible space with no monoid structure.</p> <p>Let T be a tree with the following property: </p> <p>For each point x in T, there are at least two components of T \ x which contain an at least trivalent vertex.</p> <p>In particular, the complement of x must have at least two components, so T cannot contain any 1-valent vertices.</p> <p>For example, let T be the universal cover of the theta graph.</p> <p>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.</p> <p>Such a T is homotopy equivalent to a point but can have no monoid structure.</p>