Let $latex \mu$ and $latex \nu$ are endofuncoids and $latex f$ is a funcoid from $latex \mathrm{Ob}\,\mu$ to $latex \mathrm{Ob}\,\nu$.

Then we can generalize Bourbaki’s notion of open mapping between topological spaces (that is a mapping for which images of open sets are open) by the following formula (where $latex x$ is a variable which ranges through entire $latex \mathrm{Ob}\,\mu$):

$latex V\in\langle\mu\rangle^{\ast}\{x\} \Rightarrow \langle f\rangle^{\ast}V\sqsupseteq\langle\nu\rangle\langle f\rangle^{\ast}\{x\}$.

This formula is equivalent (exercise!) to

$latex \langle f\rangle\langle\mu\rangle^{\ast}\{x\} \sqsupseteq \langle\nu\rangle\langle f\rangle^{\ast}\{x\}$.

It can be abstracted/simplified further:

$latex \mathrm{Compl}(f\circ\mu)\sqsupseteq\mathrm{Compl}(\nu\circ f)$.

The last formula looks deceitfully similar to a formula expressing continuous funcoid, but it is unrelated.

That is what open maps are in a higher abstraction level. These seem to posses no interesting properties at all (but I may mistake about this).

In my book I have shown that for co-complete funcoids being an “open” is a special case of being continuous (but being continuous this times is defined with the reversed order).