THM0009canonicalv1Categoricity of tau_0
tau_0 is categorical: any two models are uniquely isomorphic. The unique isomorphism maps rho^n(g) to (rho^M)^n(g^M).
Payload
Categoricity of tau_0
tau_0 is categorical: any two models are uniquely isomorphic. The unique isomorphism maps rho^n(g) to (rho^M)^n(g^M).
Categoricity of tau_0
Summary
tau_0 is categorical: any two models are uniquely isomorphic. The unique isomorphism maps rho^n(g) to (rho^M)^n(g^M).
Statement
%
\label{thm:categoricity}
The theory $\tau_0$ is categorical:
if $M$ and $N$ are any two models of $\tau_0$,
then there exists a unique isomorphism $\varphi : M \xrightarrow{\sim} N$.
Proof / Justification
Let $M \models \tau_0$.
We construct the unique isomorphism
$\varphi : \tau \to M$.
\medskip
\textbf{Existence.}
Define $\varphi$ by:
\begin{align*}
\varphi(\omega) &:= \omega^M, \\
\varphi(\rho^n(g)) &:= (\rho^M)^n(g^M)
\quad\text{for } g \in \{\alpha, \pi, \gamma, \eta\},\; n \geq 0.
\end{align*}
This is well-defined by the Ontic Closure Theorem
(every object of $\tau$ has a unique representation).
\emph{$\varphi$ is well-defined on $M$:}
Since $M \models \tau_0$,
$M$ satisfies $\KAxiom{3}$ (orbit-seeded generation),
so $(\rho^M)^n(g^M)$ exists in $M$ for all $n \geq 0$.
\emph{$\varphi$ preserves $\rho$:}
$\varphi(\rho(x)) = \rho^M(\varphi(x))$ by construction.
\emph{$\varphi$ preserves order:}
The order in $\tau$ is determined by
depths within and across orbits (via $\KAxiom{1}$).
The same determination holds in $M$.
\emph{$\varphi$ is injective:}
Suppose $\varphi(\rho^n(g)) = \varphi(\rho^m(h))$,
i.e., $(\rho^M)^n(g^M) = (\rho^M)^m(h^M)$.
Since $M \models \tau_0$, the orbit rays in $M$
are pairwise disjoint
(by the same proof as Proposition~\ref{prop:orbit-disjoint},
which uses only $\KAxiom{1}$--$\KAxiom{6}$).
Therefore $g = h$ and $n = m$.
\emph{$\varphi$ is surjective:}
Since $M \models \KAxiom{6}$ (Object Closure),
every element of $M$ is either $\omega^M$
or $(\rho^M)^n(g^M)$ for some $g$ and $n$.
Both are in the image of $\varphi$.
\medskip
\textbf{Uniqueness.}
Any isomorphism $\psi : \tau \to M$ must satisfy:
$\psi(g) = g^M$ (preservation of constants)
and $\psi(\rho^n(g)) = (\rho^M)^n(\psi(g)) = (\rho^M)^n(g^M)$.
Therefore $\psi = \varphi$.
Source Context
- Registry source:
book-01.jsonlline 46 - Manuscript source:
2nd-edition/book-i-categorical-foundations/02_mainmatter/part02/ch09-rigidity-categoricity.texlines 179-185
Lean / Formalization Notes
- Formalization:
formalized - Module:
TauLib.BookI.Orbit.Rigidity - Name:
Tau.Orbit.categoricity
Dependencies
- Canonical: I.T01, I.K1, I.K2, I.K3, I.K4, I.K5, I.K6, I.P01
Related Results
Generated by later projection phases.
Related Publications
Generated by later projection phases.
Revision Notes
- 2026-04-24: Initial pilot migration.
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