68 lines
2.6 KiB
Markdown
68 lines
2.6 KiB
Markdown
title: Uniquness of the absurdity proofs in Cartesian closed categories
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tags: category theory
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date: 2015-08-12 00:00
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---
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Adding the DNE to CCCs collapses them into Boolean algebras:
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Suppose $\mathcal{C}$ is a Cartesian closed category.
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We can show that if $\mathcal{C}$ has a family of isomorphisms between $A$ and $(A \to 0) \to 0$, then $\mathcal{C}$ is a poset (and a Boolean algebra).
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**Proposition**.
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For any object $A$ of $\mathcal{C}$, $A \times 0 \simeq 0$.
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*Proof*.
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${\mathit{\mathop{Hom}}}(A \times 0, B) \simeq {\mathit{\mathop{Hom}}}(0 \times A, B) \simeq {\mathit{\mathop{Hom}}}(0, B^A)
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\simeq 1$.
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Therefore, $A \times 0$ is an initial object in $\mathcal{C}$.
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Clearly $(\pi_2, \bot_{A \times 0})$ establishes the isomorphism.
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*Qed*
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**Proposition**.
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Given a morphism $f : A \to 0$, we can conclude that $A \simeq 0$.
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*Proof*.
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The isomorphism is witnessed by $(f, \bot_A) : A \to A \times 0 \simeq 0$ (where
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$\bot_A : 0 \to A$ is the unique initial morphism).
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*Qed*
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We have seen that given the morphism $f : A \to 0$ we can construct an
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iso $A \simeq 0$. In particular, that means that there can be *at
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most one arrow* between $A$ and $0$; interpreting arrows as
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"proofs" or "proof object", that means that in the intuitionistic
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calculus there is at most one proof/realizer of the negation (up to
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isomorphism).
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Now, returning to the original question, how do CCCs with
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double-negation elimination look like? Specifically, we want to show
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that if in $\mathcal{C}$ there is an isomorphism between $A$ and
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$0^{(0^A)}$ for all objects $A$, then the categorical structure
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collapses into a Boolean algebra.
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We say that a category has *double negation elimination* if for every $A$, it is the case that $A \simeq 0^{(0^A)}$.
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**Proposition**.
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Given a category with double negation elimination and two morphisms $f, g : A \to B$, it
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is the case that $f = g$.
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*Proof*.
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Let $i$ be the isomorphism between $B$ and $0^{(0^B)}$. Then,
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$i \circ f, i \circ g \in {\mathit{\mathop{Hom}}}(A, 0^{(0^B)}) \simeq {\mathit{\mathop{Hom}}}(A \times
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(0^B), 0)$.
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But the latter set can contain at most one element; thus,
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$i \circ f = i \circ g$. Because $i$ is an isomorphism, $f = g$.
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*Qed*
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It might be worth noting that in "standard" cartesian closed
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categories, a morphism $i : B \to 0^{(0^B)}$ serve as a coequalizer of
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$f$ and $g$. Perhaps stretching the analogy a bit, we can view $i$ as
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"equalizes" two constructive proofs of $B$ into a single classical
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proof.
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**See also**
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- [A categorical characterization of Boolean algebras](http://link.springer.com/article/10.1007%2FBF02485724) by M.E. Szabo
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