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Let us define complement topos as the categorical dual of a topos, with all of the arrows reversed. Remember an (elementary) Topos is defined as

a category with initial (0) and terminal (1) objects, pullbacks, pushouts, exponentiation, and a subobject classifier, which is an object Ω together with a morphism true such that for every monic m there is a unique morphism $\chi_m$ which makes the following diagram a pullback:

Diagram for a Topos

$\chi_m$ is called “the characteristic or classifying morphism of m” and ! is the unique morphism from S to 1 in C. A subobject classifier is unique up to isomorphism, and so the morphism $\chi_m$. Propositions and truth-values are morphisms $\phi: 1 \to \Omega$.

This definition comes from the paper Complement Topoi and Dual intutionistic Logic. This paper is problematic in that it tries to argue that there by relabeling true to false one gets a complement topos, but actually the ordering that a topos gives to sets is one of inclusion not the opposite, and this ordering selects true rather than false. I traced this argument to the 1995 book Inconsistent Mathematics. (It looks like the authors there, interested as they were to find the dual of a topos, came across the right concepts, then dualised it again, and tried to explain their results from the side of the topos).

Now there is a completely reasonable dual to a topos that we can get by looking at a Topos in the Opposite Category. There we will find that we still have finite limits and colimits, but instead of exponentials we will have co-exponentials (satisfying the dual property of exponentials) and a "quotient co-classifier" (an object $\mho$ equipped with an epimorphism $\bot: \mho \to 0$, where $0$ is the initial object, such that any epimorphism with domain $X$ is the pushout of $\bot$ along a unique classifying morphism $\mho \to X$)

complement topos

In order to help see what this means I came up with a simple illustration in the category of Complete Atomic Boolean Algebras (CABA) which are isomorphic to $Set^{op}$ (See Nlab page on CABA and the answers to the question What is the opposite category of Set). I am not sure yet whether $0$ should go to $\top$ as illustrated or to $\bot$.

The opposite of an object in Set is given by the powerset of that object. In Set $0$ is initial its powerset is the one element set containing the empty set which in CABA is the terminal object. In Set 1 is the terminal object and it's powerset contains two elements ${\emptyset, 1}$.

complement classifier from a 3 atom CABA to a 2 atom one

I did not add all the arrows from $\top \to \top$ and $\bot \to \bot$ as these always have to be there.

In order to deal with conjunction and disjunction I need to work out what $\mho \times \mho$ and $\mho + \mho$ look like as boolean algebra. That is I want to work out what the following look like in CABA, to check whether I have correctly dualised conjunction and disjunction.

disjunction co-classifier conjunction co-classifier

Sums and producs in CABA should be well known (and I found a few hints here and there), but I could not find clear explanations of this in my quick search on the internet, and I wanted to be really sure before proceeding.

The idea is to see if this dualisation gives some clue as to complement Heyting algebras, as the logic of falsification (where co-exponentials play an key role). We should end up with a logic of falsification.

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Well, a CABA is just a powerset. The coproduct of the powersets of $A$ and $B$ is the powerset of their product, and similarly the product is the powerset of their coproduct. That is, to construct some limit or colimit of CABAS, apply the dual construction to the sets of atoms.

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  • $\begingroup$ So usually $A + B$ is bigger than $A \times B$? I think for the case of two elements 0,1, I just drew up $A \times B$ with atoms $\overleftarrow{0}, \overleftarrow{1}, \overrightarrow{0}, \overrightarrow{1}$. A+B will then have as atoms the pairs of {(0,0),(0,1),(1,1),(0,1)} which should make for the same sized powerset. Not surprising then because left and right are encoded as 0 and 1, and $2+2=2\times2=4$. I will take up more space to draw out as pairs though... $\endgroup$ – Henry Story Apr 28 at 10:42
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    $\begingroup$ @HenryStory Yep, the coproduct is usually bigger than the product, at least in the finite case. $\endgroup$ – Kevin Arlin Apr 28 at 14:57

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