# Must diagrams be commutative?

Given a category C and a function $\Theta : Mor(\textbf{C}) \times Mor(\textbf{C}) \longrightarrow Mor(\textbf{Rel})$ and suppose that the relation, with $(r,s)\in\Theta(u,\bar{u})$ and so forth, is such that: $\require{AMScd}$ $$\begin{CD} A @>u>> B@>v>>C\\ @VrV V \Theta@VsV V\Theta @VVtV\\ \bar{A} @>>\bar{u}> \bar{B}@>>\bar{v}> \bar{C} \end{CD} \quad\Rightarrow\quad \begin{CD} A @>vu>> C\\ @VrV V\Theta @VVtV\\ \bar{A} @>>\bar{v}\bar{u}> \bar{C} \end{CD}$$ Then $\Theta$ give arise to a category with arrows as objects and pairs of arrows as morphisms. If $(r,s)\in \Theta(u,\bar{u})\Leftrightarrow s\circ u = \bar{u}\circ r$, then $\Theta$ is the commuting relation.

Since the commuting relation is so commonly used for universal definitions, one might be blind for other kinds of constructions, especially of concrete categories?

In a study of the character of the relation between constructs and their morphisms, I have in Rel discovered that the relation $(r,s)\in \Theta(u,\bar{u})\Leftrightarrow [(a,b)\in u\wedge (\bar{a},\bar{b})\in \bar{u}\wedge (a,\bar{a})\in r \Rightarrow (b,\bar{b})\in s]$, seems to satisfy my needs (even if it perhaps not seems so sexy). But when I tried to present this idea, an obviously skillful mathematician drew the conclusion that I must have meant the relation $(r,s)\in \Theta(u,\bar{u})\Leftrightarrow s=\bar{u}\circ r\circ u^{op}$, since this latter $\Theta$ implied the the previous original $\Theta$.

1. Are the two definitions equal?
2. Does successful ideas have too heavy impact on our thinking?
3. Is there room for alternative conditions on diagrams in category theory?

(I admit that it took years before I realized that it perhaps shouldn't be commutative diagrams, neither with nor without reversed arrows...)

The idea is presented in MO

• In $2$-category, almost nothing is commutative. – Jakob Werner Aug 31 '14 at 7:16