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For sets $X,Y$ and a function $f: X\times Y\to X$, what is the name of the property whereby for all $x\in X$ and $y_1, y_2 \in Y,$ $$f(f(x,y_1), y_2) = f(f(x, y_2),y_1)\qquad?$$

Some of us called it a "generalized associativity," some "generalized commutativity," some a combination of both.

Anything we can find on associativity and commutativity and so on assumes the binary relation is inside some $X$.

We used this inside of a proof, but can't find the analogous mathematical concept.

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    $\begingroup$ Mathematicians would write the question thus: if $f:X\times Y\to X$ then what is the name of the property $f(f(x,y_1),y_2)=f(f(x,y_2),y_1)$? If you think of $f$ as giving a right action of $Y$ on $X$, i.e. a map from the set $Y$ to the semigroup of endomorphisms of $X$, then the assertion is just that the image of $Y$ lands in a commutative subsemigroup. $\endgroup$
    – eric
    Oct 12, 2015 at 20:40
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    $\begingroup$ Said another way, you have a family of endomaps $f_y : X \to X$ parameterized by $y \in Y$, and the desired condition is that these maps commute. $\endgroup$
    – Qiaochu Yuan
    Oct 12, 2015 at 22:29
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    $\begingroup$ The two comments above nicely illustrate an peculiarity of notational taste I’ve always found somewhat amusing. Most mathematicians dislike the curried notation $f : A \to (B \to C)$, and strongly prefer the form $f : (A \times B) \to C$. But the very same people may be quite happy to say things like “a family of maps $f_a : B \to C$, parametrised by $a \in A$”. It seems to be some sort of residual aversion to the explicit consideration of higher-order types and functionals. $\endgroup$
    – Peter LeFanu Lumsdaine
    Oct 13, 2015 at 11:41
  • $\begingroup$ Even with some Google assistance I've been unable to decipher the meaning of the second sentence ("PL-group" / "professional reject"). $\endgroup$
    – YCor
    Dec 5, 2016 at 6:36
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    $\begingroup$ Speaking informally, I'm hard-pressed to see anything associative in this; any way you slice it, it's not simply a 'regrouping' of terms but very clearly involves a reordering of operations among the $y$s. It is much more commutative than associative in flavor (and this can be formalized, of course, as in the answer given). $\endgroup$
    – Steven Stadnicki
    Dec 5, 2016 at 7:23

1 Answer 1

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You could say "$f$ makes $X$ into a commutative $Y$-unary algebra" or that "$f$ is a commutative $Y$-unary algebra structure on $X$."

For instance, defining

$$f : \{N,E\} \times \mathbb{N}^2 \rightarrow \mathbb{N}^2$$ by $$f(N,(x,y)) = (x+1,y) \qquad f(E,(x,y)) = (x,y+1)$$

we have that $(\mathbb{N}^2,f)$ is the commutative $\{N,E\}$-unary algebra freely generated by $\{(0,0)\}$.

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