Just a few minutes ago I’ve formulated a new important conjecture about funcoids:

Let $latex A$, $latex B$ be sets.

**Conjecture** Funcoids $latex f$ from $latex A$ to $latex B$ bijectively corresponds to the sets $latex R$ of pairs

$latex (\mathcal{X}; \mathcal{Y})$ of filters (on $latex A$ and $latex B$ correspondingly) that

- $latex R$ is nonempty.
- $latex R$ is a lower set.
- $latex R$ (ordered pointwise) is a dcpo

by the mutually inverse formulas:

$latex (\mathcal{X} ; \mathcal{Y}) \in R \Leftrightarrow \mathcal{X} \times^{\mathsf{FCD}} \mathcal{Y} \sqsubseteq f \quad \mathrm{and} \quad f = \bigsqcup^{\mathsf{FCD}} \left\{ \mathcal{X} \times^{\mathsf{FCD}} \mathcal{Y} \mid (\mathcal{X} ; \mathcal{Y}) \in R \right\}$.

Oh, a trivial counter-example: $latex f = ([0 ; 2] \times [0 ; 1]) \cup ([0 ; 1] \times [0 ; 2])$. I will try to figure another axioms.

Oh, sorry, the counter-example is wrong. However it inspired me with another similar conjecture.

No, it is wrong. For a counter-example take $latex R=\{\uparrow\emptyset,\uparrow\{0\},\uparrow\{1\}\}$.