Let $(B,\ast)$ be a magma (that is, $\ast:B\times B\to B$ is a binary operation on $B$), and let $\tau$ be a topology on $B$. If $X$ is any set and we define $\tilde\ast$ in the set $B^X$ of functions from $X$ to $B$ componentwise (that is, if $f,g:X\to B$, we define $f\,\tilde\ast\,g:X\to B$ via $(f\ast g)(x)=f(x)\ast g(x)$, for each $x\in X$), then $(B^X,\tilde\ast)$ is also a magma.

If $\ast$ is continuous (where we endow $B\times B$ with the product topology), we say that $(B,\ast,\tau)$ is a topological magma. In such case, for any topological space $X$, the subset $\mathcal C(X,B)$ of continuous functions from $X$ to $B$ is a submagma of $B^X$ (because for $f,g$ as before we have that $f\,\tilde\ast\,g$ is the composite map $x\mapsto\bigl(f(x),g(x)\bigr)\mapsto f(x)\ast g(x)$). I was wondering if the converse is true:

If $(B,\ast,\tau)$ is such that, for every topological space $X$, the set $\mathcal C(X,B)$ is a submagma of $(B^X,\tilde\ast)$, does it follow that $(B,\ast,\tau)$ is a topological magma?

  • $\begingroup$ subring should be submagma, I suppose? $\endgroup$ – Henno Brandsma Sep 24 '17 at 5:27
  • $\begingroup$ @HennoBrandsma You are right! correcting now. $\endgroup$ – Matemáticos Chibchas Sep 24 '17 at 6:01

The projection maps $\pi_1,\pi_2\in C(B\times B,B)$ satisfy $\pi_1\,\tilde\ast\,\pi_2=\ast$. So $(B,\ast,\tau)$ is a topological magma if and only if $\mathcal C(B\times B,B)$ is closed under $\tilde\ast$.


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