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September 1, 2013 / porton

Open mappings between endo-funcoids

Let \mu and \nu are endofuncoids and f is a funcoid from \mathrm{Ob}\,\mu to \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 x is a variable which ranges through entire \mathrm{Ob}\,\mu):

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

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

It can be abstracted/simplified further:

\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).


One Comment

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  1. porton / Sep 2 2013 18:44

    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).

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