[personal profile] sassa_nf 2020-06-29 06:48 am (UTC)(link)
Can you add something about Χf? It is clear from the examples it is not an arbitrary arrow B→Ω.

Well, ok, maybe this is a stupid question. Of course, it's just an arrow that makes the square commute.
Edited 2020-06-29 10:29 (UTC)

[personal profile] sassa_nf 2020-06-29 02:41 pm (UTC)(link)
Chi_f in the diagram for the definition.

epimorphisms_split: (Default)

[personal profile] epimorphisms_split 2020-06-29 06:49 am (UTC)(link)
картинки маленькие, можно увеличить раза в два.

[personal profile] sassa_nf 2020-06-29 10:16 am (UTC)(link)
Also, is the arrow Ω1→Ω⨯Ω labelled (true,true) correctly?

[personal profile] sassa_nf 2020-06-29 10:26 am (UTC)(link)
Is the introduction of Union of Subobjects necessary to define disjunction? Can we not use ∨: Ω⨯Ω→Ω = ⇒ . (¬⨯id)?
Edited 2020-06-29 10:27 (UTC)

[personal profile] sassa_nf 2020-06-30 03:02 pm (UTC)(link)
I am not sure what the problem with double-negation is, if everything is reduced to true/false anyway.

[personal profile] sassa_nf 2020-06-30 05:33 pm (UTC)(link)
Hmmm.... I thought Ω was an object with just two points.

But then if Ω can have more than two points in some logic, then the definition of ∧ does not seem complete, because it doesn't explain how to choose mappings other than (true,true)→true.

If it's only two points in Ω, there is no choice; but if it's more points, then the specified definition doesn't seem to make ∧ a unique arrow.

[personal profile] sassa_nf 2020-06-30 05:49 pm (UTC)(link)
ok, but what about Heyting algebra. Does this mean Ω = {a, b, T, ⊥} ? How does ∧ know what a ∧ b is, since only T∧T is specified?

[personal profile] sassa_nf 2020-06-30 06:04 pm (UTC)(link)
Heyting algebra: a lattice of ⊥ < a < T and ⊥ < b < T. Then x ∧ y = max({c for c in Ω and c ≤ a and c ≤ b}).

But my question is how does the definition reflect this fact? It seems to only say that T ∧ T = T - nothing about a ∧ b = ⊥, and not, say, a ∧ b = a.
Edited 2020-06-30 18:06 (UTC)

[personal profile] sassa_nf 2020-06-30 08:17 pm (UTC)(link)
Sure.

Maybe I don't get something basic. What it looks like to me, since the square has arrows (true,true) and true, it defines only the behaviour of the classifying arrow for one pair, not all possible pairs.