Visual\Geometric characterization of associativity Given a group $(G,f)$, I'm trying to characterize the graph of $f$ maybe in a more visual or geometric way. I'll explain a bit more:
1) because $f$ admits the zero element axiom of a group operation, we must have:
$f$ restricted to the line $x=0$ is the identity function on $M$, and same for the line $y=0$. (we can also replace zero by other unique element) 
2) because $f$ admits the inverse axiom of a group operation, we must have:
$f$ restricted to every line $x=a$ is a bijection $M \to M$ (quiet easy to see).
My question is, can you find a similar characterization for the third axiom, the associativity axiom. all I can get so far is that $f(f(a,b),c) = f(a,f(b,c))$), and that's  the definition. Maybe I'm looking for a small insight which is more visual, geometric, or  just more developed than the definition. Perhaps something about the structure of the restrictions of this function to its fibers, or something else that I don't see and might help.
Will appreciate your help! 
 A: Shasheff polytopes aka Associahedra are geometric representations to study associativity, after Jim Stasheff who first studied them in the 1960s.

The associahedron Kn is a polytope of dimension n whose vertices are
  in one to one correspondence with the parenthesizings of the word x0x1
  . . . xn+1.


From the 2005 introduction by Loday "The multiple facets of the associahedron", which contains a compact history of the subject and summarizes links between algebra, geometry and topology of associativity. Loday goes on to treat associativity in terms of vector spaces of geometric binary trees.
A: I'm all for exlaining math in avisual way, but in this case I'm not sure you're going to succeed. On the one hand, explaining associativity requires taking a result (i.e. a $z$ value of your plot) and feeding it back as an input (an $x$ or $y$ value). While this can be explained in terms of some obscure projection sequence, this is hardly intuitive.
On the other hand, I don't fully agree with what you have so far. For example over the reals, the function $f:(x,y)\mapsto\frac{x+y}2$ is bijective in every restriction, but there still is no universal inverse for every element. I know this function violates the first axiom, but even I'm not sure just now whether your form of the first two axioms actually implies the common first two axioms.
