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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.

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  • $\begingroup$ I'm curious if the standard model of $\mathbb {CP}^\infty$ admits a monoid structure. It's homotopy equivalent to a monoid but I don't know how to put a monoid structure on it directly. $\endgroup$
    – Vitali Kapovitch
    Commented Mar 15, 2012 at 23:03
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    $\begingroup$ @Tom Goodwillie: But if $xy \equiv 1$ then $1$ must be the monoid's identity element, so $x \equiv x1 \equiv 1$ and the space is a singleton. $\endgroup$
    – Noah Stein
    Commented Mar 15, 2012 at 23:43
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    $\begingroup$ @Vitali: If you view $\mathbb{CP}^\infty$ as the space of nonzero polynomials $f(x)$ mod the equivalence relation $f(x) \sim \lambda f(x)$, you obtain a topological monoid structure. This is essentially the same as the identification of $\mathbb{CP}^\infty$ with the infinite symmetric product of $S^2$. Whether you can make it a topological group is more interesting. (I've heard a construction of it as a topological group using units in $\mathbb{C}(x)$, but this secretly replaces $\infty$ with the continuum.) $\endgroup$
    – Tyler Lawson
    Commented Mar 16, 2012 at 0:57
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    $\begingroup$ What are some necessary conditions for a space to admit a topological monoid structure? Having abelian fundamental group is one, but that's useless here because it's homotopy-invariant. So, what non-homotopy-invariant necessary conditions are there? $\endgroup$
    – Tom Leinster
    Commented Mar 16, 2012 at 6:29
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    $\begingroup$ @Tom The fundamental group only needs to be abelian at base points corresponding to units of $\pi_0$. For example let A be any space. Then we can make the space $\{0 \} \cup A \cup \{\infty\}$ into a monoid by declaring 0 to be the unit and that the product of any two non-identity elements is $\infty$. This example also shows that it will be very hard to get even homotopy-invariant necessary conditions, let alone more refined invariants. $\endgroup$
    – Chris Schommer-Pries
    Commented Mar 16, 2012 at 13:12

4 Answers 4

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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.

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    $\begingroup$ I played fast and loose with 2-sided invertibility, but it doesn't matter. Either: -there is a point which is not invertible. The argument goes through as above. -every element is invertible on one side or the other. Assume x is left invertible but not right invertible, and y is the reverse. xy is WLOG left invertible or right invertible so y is left invertible, a contradiction. Then WLOG every left invertible element is 2-sided invertible, so every right invertible element is, so every point in the monoid is 2-sided invertible. $\endgroup$
    – Gabriel C. Drummond-Cole
    Commented Mar 21, 2012 at 17:50
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    $\begingroup$ Hi Gabriel, very nice answer, thanks! Let me summarize and simplify a bit your argument: Let M be a monoid; then: i) for each pair x,y of invertible elements there exist a homeomorphism of M in itself mapping x to y; ii) either M is a group or there exist an element which has neither a left inverse nor a right inverse; iii) if M is path connected but M-{e} is not connected then all elements of M except at most those in one connected component of M-{e} are invertible. Thus a monoid structure on the infinite ternary graph would give a homeomorphism mapping an edge internal point onto a vertex. $\endgroup$
    – domenico fiorenza
    Commented Mar 23, 2012 at 9:39
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Let E be a contractible space, and $X = E \coprod \{0\}$. 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 $A_\infty$ structure on $X$.

Suppose we could make this come from a topological monoid structure. By checking the induced monoid structure on $\pi_0$, 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.

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.

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.)

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    $\begingroup$ This doesn't quite answer the original question. He wanted an example with no topological monoid structure at all. The reals with addition is a contractible topological group and you can add a new isolated point which is an absorbing element and it is a topological monoid of the above sort. But course it doesn't induce the Z/2 structure. $\endgroup$
    – Benjamin Steinberg
    Commented Mar 16, 2012 at 1:18
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    $\begingroup$ While this doesn't give a space X with has no monoid structre at all, it does give a very elementary example where there is no monoid structure compatible with a given monoid structure on $\pi_0$. This avoids all but the simplest parts of the "equivalence issue" that Domenico was alluding to. It seems to me to be a satisfactory answer to the question if no one can produce such a space X. $\endgroup$
    – Chris Schommer-Pries
    Commented Mar 16, 2012 at 1:59
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    $\begingroup$ Thanks, Tyler! That example perfectly suits me! For my teaching purposes it is even better than a space with no topological monoid structure at all, since it allows me to stress on the importance of the fact that a homotopy equivalence of topological spaces is promoted to a equivalence of $A_\infty$ spaces. I'm temporarily leaving the question open in its original form (adding a few comments) as an (admittedly odd) question in general topology. $\endgroup$
    – domenico fiorenza
    Commented Mar 16, 2012 at 6:47
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See here: Algebras over the little disks operad

The question asked there was different/more general, but the answers actually address the restricted question asked here.

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  • $\begingroup$ Hi Neil, thanks for pointing my attention to that. But I'm not sure that completely solves the question, yet: it seems to me the answers there suggest that if such a space X does exist, then one of the simplest examples of such a space is likely to be a tree; but it seems that the question whether there actually exist trees with no admissible monoid structure is left open. Or am I missing something? $\endgroup$
    – domenico fiorenza
    Commented Mar 16, 2012 at 8:37
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    $\begingroup$ As it turns out, every tree has a monoid structure. Privilege one monovalent vertex as 1, and identify the edge incident on that vertex as [0,1]. So we can write the tree as [0,1]\cup T, where T is a tree which intersects [0,1] at the point 0. Give each edge of T length 1. There is a canonical path from any point outside [0,1] to 0, which has now been given a length. Let [0,1] act on itself by multiplication, let T * T = 0, and for c in [0,1] and x in T, let cx=xc be the point a distance of c * d(x,0) from 0 along the path between 0 and x. $\endgroup$
    – Gabriel C. Drummond-Cole
    Commented Mar 16, 2012 at 17:28
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    $\begingroup$ I was thinking of finite trees; I guess this works as long as there is a 1-valent vertex and as long as every point in the tree is a finite number of edges away from that vertex. So there still could be some badly infinite tree or a countably infinite tree with no boundary as a counterexample. $\endgroup$
    – Gabriel C. Drummond-Cole
    Commented Mar 16, 2012 at 18:55
  • $\begingroup$ And there is... $\endgroup$
    – Gabriel C. Drummond-Cole
    Commented Mar 21, 2012 at 10:43
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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.

Edit. It is shown here that no monoid structure can be put on this space for which multiplication is separately continuous.

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  • $\begingroup$ The link to springerlink.com is broken. I'm also unable to find any snapshot saved on the Wayback Machine. $\endgroup$
    – The Amplitwist
    Commented Apr 16, 2023 at 15:33
  • $\begingroup$ @TheAmplitwist, this was more than 10 years ago. I don't recall what the link pointed at $\endgroup$
    – Benjamin Steinberg
    Commented Apr 16, 2023 at 17:40
  • $\begingroup$ Quite understandable. Thanks for taking a look all the same :) $\endgroup$
    – The Amplitwist
    Commented Apr 16, 2023 at 18:33

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