# Is this a general structure for constructs?

Here a construct is a category where the objects are sets and the morphisms are structure preserving functions. Common examples are groups, graphs and topological spaces. As far as I can see there is a very general structure for all constructs:

A simple $F$-construct is a set and a binary relation $(X,\mathcal R)$ such that $\mathcal R\subset F(X)\times X$. The functor $F$ characterize the primary structure that determines the conditions on the morphisms and there could be secondary axioms on $\mathcal R$ separating categories from each other. More general constructs can be obtained by combining simple constructs.

For example, if $F(X)=X\times X$, then $(X,\mathcal R)$ could be anything from categories
$(X=\text{Mor}(\underline C)$ with $((\alpha,\beta),\gamma)\in\mathcal R \iff \beta\alpha=\gamma)$ to groups when $\mathcal R$ is an associative function etc.

Non algebraic examples are topological spaces

$F(X)=\mathcal P(X)$ (power set) with $\mathcal (A,x)\in \mathcal R\iff x\in \bar A$

What I'm looking for is possible counterexamples, known and newly found constructs, that doesn't seems to fit in to the pattern.

As Tobias Kildetoft pointed out in a comment a relation is a very general approach $-$ and that is intentional. But $F$-constructs is about a normal form for all constructs as defined above and given in this normal form there is a general rule that determine the condition for a function to be a morphism.

A morphism is a function $f:X\rightarrow Y$ satisfying the relation:

$(1)\quad\; (\phi,\psi)\in F(f)\Rightarrow\left[(\phi,x)\in\mathcal R^F_X\Rightarrow(\psi,f(x))\in\mathcal R^F_Y\right]$.

The meaning with the construction of the relation is demonstrated by the diagram: $\require{cancel}$ $\require{AMScd}$ \begin{CD} F(X) @>F(f)>> F(Y)\\ @V \mathcal R_X^F VV\xcancel{\#} @VV\mathcal R_Y^F V\\ X @>>f> Y \end{CD}

The upper row gives a condition on $f$ on the lower row, together with the relations, and the connection is given by $(1)$. (The diagram commutes if the $\mathcal R$:s are functions and $F(f)$ is a function or a anti-function.

Of course, if someone find a counterexample of this condition, I would be interested to know that too.

Examples of how $(1)$ works could be find here.

I really don't understand the resistance against the question. The method works, and it is not easy to find adequate counterexamples.

• Your definition of a construct does not seem to really pose any restriction to an arbitrary category. Is there some requirement that the arrows actually be maps of the sets? – Tobias Kildetoft Nov 27 '14 at 12:58
• @Tobias Kildetoft: The condition is the relation, a sort of normal form. And yes, the arrows should be functions. – Lehs Nov 27 '14 at 13:04
• As the last time... Please try to be more rigorous when you write things. "A construct is a category where the objects are sets": everything in (standard) mathematics is sets. Based on your comments, do you mean to say that your category is equipped with a forgetful (= faithful) functor to $\mathsf{Set}$? Is $F$ a functor $\mathsf{Set} \to \mathsf{Set}$? Etc. – Najib Idrissi Nov 27 '14 at 13:09
• @Lehs: You have already the same question a couple of times on math.SE and I had already given you counterexamples (sheaves for example), and indicated that there are tons and tons of counterexamples, and that therefore your "universal approach" does not make much sense. – Martin Brandenburg Nov 27 '14 at 14:34
• @Lehs, you're welcome. I don't know much about coalgebras, but here's a link to get you started. Don't worry too much about the resistance; most great ideas were resisted initially. Anyway, the important thing is not to get 100% of all kinds of mathematical structures ever considered. The important thing is whether it always gets the morphisms right. So I think you should work out, in detail, what the morphisms are in a large variety of cases. If your method consistently gets the right answer, then consider publishing an article. – goblin Nov 27 '14 at 19:58