Comparing "axiomatized function spaces"

Question

This was previously asked and bountied at math.stackexchange with no response. I've also tweaked the language for clarity; see the edit history for the broader context, and note that the existing answers were aimed at the earlier version of the question.

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Let $C(\mathbb{R}^2,\mathbb{R})$ be the space of all continuous functions $\mathbb{R}^2\rightarrow\mathbb{R}$ with the compact-open topology, and consider the following two subspaces:

• $\mathcal{Asso}$ = the subspace of all associative functions.

• $\mathcal{Comm}$ = the subspace of all commutative functions.

Of course these are distinct as subsets; my question is whether they are topologically distinguishable:

Is $\mathcal{Asso}\cong\mathcal{Comm}$?

I strongly suspect that the answer is negative - in particular, $\mathcal{Comm}$ is connected and I suspect $\mathcal{Asso}$ is not - but I don't see how to prove this. (Another nice feature of $\mathcal{Comm}$ is that it is an $\mathbb{R}$-vector space in a natural way; however, it's not clear how to extract any power from this observation.)

Generally, $\mathcal{Asso}$ seems rather mysterious, and I'm interested in any information about it.

Answer 1

The following result from Section 6.2 of János Aczél's classic book Lectures on Functional Equations and their Applications seems to be closely related, and the long list of references on the solutions of the associativity equation (p.253/4) may provide further entry points into the literature. (Although that list is way out of date, with the book being from 1966.)

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Theorem: Let $f \in C(\mathbb{R}^2,\mathbb{R})$ satisfy the associativity equation $$f(x,f(y,z)) = f(f(x,y),z).$$ Suppose moreover that $f$ is cancellative, i.e. that $$ f(x,y) = f(x,z) \quad \Rightarrow \quad y = z, $$ and similarly in the other argument. Then $f$ is conjugate to addition $+ : \mathbb{R}^2 \to \mathbb{R}$ via a monotone homeomorphism. In particular, $f$ is commutative.

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It seems plausible that this result can be used to prove that if your theory $T$ consists of associativity and the cancellativity laws, then $T_{\mathbb{R}}$ is homemorphic to the space of monotone homeomorphisms $\mathbb{R} \to \mathbb{R}$. But I haven't worked out the details.

Answer 2

Here are some functions in the connected component of the associative function $f(x,y)=x$.

Proposition: If $f(x,y)=g(x)$ and $f$ is associative, then there is an interval $[a,b]$ (possibly infinite on either side) such that $$g(x)=x \text{ for all }x\in [a,b]$$ $$g(x)\in[a,b]\text{ for all }x\in \mathbf{R}$$ Furthermore all such $g$ produce associative $f$.

Proof: By associativity, $g(g(x))=g(x)$. So if $y$ is in the range of $g$ then $g(y)=y$. Then let $a=\inf(range(g))$ and let $b=\sup(range(g))$. The other direction is trivial. $\square$

So in any case, the connected component of $f(x,y)=x$ is substantial. As an easy example, the function $f(x,y)=x$ is the limit of $f_n(x,y)=\max(n,x)$.

It seems plausible to me that these associative functions are disconnected from the ones that depend on both variables. If so, the associative subspace would be disconnected.