Abstract formulation of associativity

Question

Say we are given a binary operation $f$ on a set $X$, that is, $$ f : X \times X \to X. $$ Denote by $\text{Id}$ the identity map on $X$. We say that $f$ is associative if, for all $x, y, z \in X$, we have $$ f(f(x, y), z) = f(x, f(y, z)). $$ I was wondering if there is a more abstract way of formulating this relationship, i.e. in a coordinate-free way. After looking at a couple of commutative diagrams, I came up with the following: $$ f \circ (f \otimes \text{Id}) = f \circ (\text{Id} \otimes f), $$ where $\otimes$ is defined through $$ (f \otimes g)(x, y) = f(x)g(y). $$ Is this a senisble abstract definition of associativity? Can it be simplified somehow? My goal ultimatiely us to understand associativity as a "form of higher-order commutativity", if that makes sense. Am I onto something here?

Answer 1

I assume you meant $(f\otimes g)(x,y)=\big(f(x),g(y)\big)$. Yes, it is sensible. In fact this can be generalized to any category with products. And so we can even get rid of pointwise definition, as long as you know how to form products. In particular that's how group objects are defined over any category.

Answer 2

My goal ultimatiely us to understand associativity as a "form of higher-order commutativity"

Indeed, if $L_x$ is left "multiplication" with $x$ (that is, $L_x(y) = f(x,y)$) and $R_y$ is right multiplication with $y$, then $f$ is associative iff every $L_x$ comutes with every $R_y$.