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Commit fea7db8c authored by Max New's avatar Max New
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define free freyd category, conjecture about preorder version

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\usepackage{amsmath,amssymb}
\usepackage{tikz-cd}
\usepackage{mathpartir}
\newtheorem{theorem}{Theorem}
\newtheorem{definition}{Definition}
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\newcommand{\relto}{\to}
\newcommand{\M}{\mathcal{M}}
\newcommand{\sq}{\square}
\newcommand{\lett}{\text{let}\,\,}
\begin{document}
\title{Freyd Multicategories as Generalized Multicategories}
......@@ -96,19 +97,139 @@ C_l \times D_t \arrow[r] & (C \sq D)_l
\end{enumerate}
I checked on the board that it maps 2-cells, and it must be
functorial, right?
The unit of the funny tensor product is the terminal object: $1 \to
1$.
\end{definition}
By abstract nonsense, the funny tensor product on the equipment of
matrices becomes a monoidal product on the equipment of categories,
and then a planar premonoidal category is a monoid object wrt the
funny tensor product. In more pedestrian terms, a planar premonoidal
and then a (planar) premonoidal category is a monoid object wrt the
funny tensor product. In more pedestrian terms, a (planar) premonoidal
category has a monoidal category of tight morphisms, a premonoidal
category of loose morphisms and all tight morphisms are central.
A freyd category is then a \emph{cartesian} premonoidal category:
i.e. a premonoidal category where the monoidal structure on the tight
morphisms is a cartesian product. I don't know how to describe this in
terms of adjoints.
i.e. a premonoidal category where the monoidal structure on the
\emph{tight} morphisms is a cartesian product. I don't know how to
describe this as an adjoint or something in the category of sierpinski
categories, but you can just use the functor to ordinary categories
that takes the tight category and say the monoidal structure is
cartesian.
\subsection{(Strict) Freyd Categories are Monadic}
Next, we show that Freyd categories can be presented as algebras of a
monad.
%
This is also true of planar and symmetric premonoidal categories, but
we focus on the cartesian case here.
\begin{definition}[Free Strict Freyd Category]
We define the free strict Freyd category $TC$ of a Sierpinski
category $C$ as follows.
\begin{enumerate}
\item The objects are lists of objects of $C$: $(TC)_0 = C_0^*$.
\item The tight morphisms are given by the free cartesian category,
i.e. they are \emph{substitutions}:
\[
TC(a_1,\ldots,a_n;b_1,\ldots,b_m)_t = \sum_{\rho : [m] \to [n]}\prod_{0 < i \leq m} C(a_{\rho i},b_i)
\]
\item The loose morphisms are defined inductively as follows.
\begin{enumerate}
\item First we have the inclusion of tight morphisms: for every
substitution $\gamma : TC(\Gamma,\Delta)_t$ there is a loose
morphism $i(\gamma) : TC(\Gamma,\Delta)_l$.
\item Next, we need to include the loose generators, so for every
$f : C(a,a')_t$, we have a rule
\begin{mathpar}
\inferrule
{f : C_t(a_i,b)\and
p : TC(a_1,\ldots,a_n;\Gamma)
}
{\lett n+1 = f(i); p : TC(a_1,\ldots,a_n,b;\Gamma)}
\end{mathpar}
\item Next, we define composition in two steps. First, we define
composition of a loose morphism with a tight morphism by
induction on the (output) loose morphism. Then we define
composition of loose morphisms by induction on the input
morphism.
\begin{mathpar}
\begin{array}{rcl}
(\lett n+1 = f(i);p) \circ (\rho, \gamma : TC(\Gamma, a_1,\ldots,a_n)) & = &
\lett |\Gamma|+1 = (f \circ \gamma(i))(\rho(i)); (p \circ (\rho,\gamma)[|\Gamma|+1/|\Gamma|+1])\\
(\rho,\gamma) \circ (\rho',\gamma') & = & \cdots
\end{array}
\begin{array}{rcl}
(\lett n+1 = f(i);p);q &=& \lett n+1 = f(i); (p;q)\\
(\rho,\gamma);q &=& q \circ (\rho,\gamma)
\end{array}
\end{mathpar}
\item First, there are two ways to include a tight generator: in a
substitution and using the inclusion of loose generators, so we
add an axiom that those are equal:
\[
(\lett 2 = i(v)(1); (1,2)) \cong (1,v(1))
\]
which we can write in a more reduction rule style:
\[
(\lett n+1 = i(v)(j); p) \cong p \circ v(j)/n+1
\]
\item Finally, to ensure functoriality of the inclusion of loose
generators, we add in a functoriality axiom.
\[
(\lett 2 = f(1); \lett 3 = g(2); (3)) \cong (\lett 2 = (g \circ f)(1); (2))
\]
\end{enumerate}
\end{enumerate}
\end{definition}
I claim that that's a monad, its strict algebras are strict freyd
categories and its pseudo-algebras are freyd categories.
\section{Freyd Multicategories from the Free Freyd Category Monad}
A Freyd Multicategory is then a normalized T-monoid, following
Crutwell-Shulman.
%
Instead, we will use \emph{discrete} T-monoids, those whose underlying
object category is a set, though I don't think the theorem in
Crutwell-Shulman here applies.
\begin{definition}[Freyd Multicategory]
A Freyd Multicategory $M$ consists of
\begin{enumerate}
\item A set of objects $M_0$
\item For every object $A \in M_0$ and list of objects $\Gamma \in
M_0^*$, a set of tight morphisms $M(\Gamma;A)_t$ and loose
morphisms $M(\Gamma;A)_l$.
\item Tight composition
\item Loose composition
\end{enumerate}
\end{definition}
\subsection{Free Freyd Multicategory}
\section{Preordered Freyd Categories and Multicategories}
Now we want to combine Freyd (multi)categories with preordered
categories to get preorderd Freyd (multi)categories.
%
The difficulty is that preordered categories are a type of
\emph{internal} category, but Freyd categories are a type of
\emph{enriched} category.
Thinking heuristically, we probably want our version of a Mono
preorder category to be equivalent to a full embedding of preorder
categories: there are types, a preorder category modeling effectful
terms and ordinary type precision, and a wide (and tall?)
sub-double-category of pure terms and pure type precision.
Conjecture: a Mono preorder category is a preorder internal to Mono
categories.
\end{document}
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