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Given any function $f$, I'm interested in knowing whether one could always define an operator $\circ$ (that is, a two place function) such that $$f(a\circ b)=f(a)\circ f(b)$$.

Now, if there exists some element $c$ in the domain of $f$ such that $f(c)=c$ then one could set $a\circ b=c$ for all $a, b$. Yet this is quite trivial. I was wondering if such an operator could always be defined in some non-trivial manner.

Eric Wofsey
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Sam
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    You seem to be looking for homorphisms. –  Jul 04 '19 at 17:44
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    So you want the domain and the codomain of $f$ to be the same, say $X$ and a "operation" $\circ, : X \times X \to X$ ? – Carlos Esparza Jul 04 '19 at 17:48
  • One problem is where $\circ$ is defined: What does $a\circ b$ mean and $f(a)\circ f(b)$? If $f:A\to A$ and $(A,\circ)$ is some algebraic structure with respect to $\circ$, then $f$ is just a homomorphism with respect to $\circ$. – I was suspended for talking Jul 04 '19 at 17:48
  • I think what the OP is trying to say is that given some map of sets $f : S \to T$, can we always define some symbolic operation $\circ$ on $S$ and $$ on $T$ such that $f: S \to T$ becomes a morphism of monoids $(S,\circ)$ and $(T,)$. – Ruben Jul 04 '19 at 17:51
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    @Ruben As far as I can see OP is talking about magmas, not monoids. – Carlos Esparza Jul 04 '19 at 17:52
  • @0x539 Right, I had assumed too much. – Ruben Jul 04 '19 at 17:55

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Well, I don't know what you consider "trivial", but such a binary operation does always exist: you can always use the binary operation $x\circ y=x$. Then $f(a\circ b)=f(a)=f(a)\circ f(b)$. Similarly, $x\circ y=y$ would also work.

Note moreover that given any such binary operation $\circ$, another such binary operation is $x*y=f(x\circ y)$, since $f(a*b)=f(f(a\circ b))=f(f(a)\circ f(b))=f(a)*f(b)$.

In general, there may not be any examples besides these (that is, every example comes from either $x\circ y=x$ or $x\circ y=y$ by composing with $f$ some number of times). This is true rather vacuously if $f:X\to X$ where $X$ has at most one element. Less vacuously, if $f:\{0,1\}\to\{0,1\}$ is given by $f(0)=1$, $f(1)=0$, some simple casework shows that the only binary operations that work are $x\circ y=x$, $x\circ y=f(x)$, $x\circ y=y$, and $x\circ y=f(y)$. (Sketch of proof: composing with $f$ if necessary, we may assume $0\circ 0=0$, which then implies $1\circ 1=1$. Then by symmetry we may assume $0\circ 1=0$, which implies $1\circ 0=1$ and so we have $x\circ y=x$.)

Eric Wofsey
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