2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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 $\tilde\ast$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.