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14
doc/cubical.tex
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@ -11,8 +11,8 @@ normal definition of judgmental equality is an inductive data type. Cubical Agda
discards this type in favor of a new primitives that has certain computational
properties exclusive to it.
Exceprts of the source code relevant to this section can be found in appendix
\S\ref{sec:app-cubical}.
The source code can be browsed online and is linked in the beginning
of \S\ref{ch:implementation}.
\subsection{The equality type}
The usual notion of judgmental equality says that given a type $A \tp \MCU$ and
@ -137,11 +137,13 @@ With this we can now prove the desired equality $f \equiv g$ from section
%
\begin{align*}
p & \tp f \equiv g \\
p & \defeq \funExt\ \lambda n \to \refl
p & \defeq \funExt\ \phi
\end{align*}
%
Paths have some other important properties, but they are not the focus of
this thesis. \TODO{Refer the reader somewhere for more info.}
Here $\phi \tp \prod_{n \tp \bN} \var{zeroLeft}\ n \equiv
\var{zeroRight} n$. Paths have some other important properties, but
they are not the focus of this thesis. \TODO{Refer the reader
somewhere for more info.}
%
\section{Homotopy levels}
In ITT all equality proofs are identical (in a closed context). This means that,
@ -273,7 +275,7 @@ $$
We have the function:
%
\begin{equation}
\pathJ\ D\ d \tp \prod_{b \tp A} \prod_{p \tp a ≡ b} D\ a\ p
\pathJ\ D\ d \tp \prod_{b \tp A} \prod_{p \tp a ≡ b} D\ b\ p
\end{equation}
%
A simple application of $\pathJ$ is for proving that $\var{sym}$ is an

170
doc/implementation.tex
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@ -1,12 +1,27 @@
\chapter{Category Theory}
\label{ch:implementation}
This implementaiton, including this report, is available as open
source software at:
%
\begin{center}
\gitlink
\end{center}
%
All modules imported for this formalization can be browsed at this
link\footnote{%
In case the linked sources are unavailable the html
documentation can be generated by navigating to the root directory
of the project and executing \texttt{make html}.%
}:
%
\begin{center}
\doclink
\end{center}
This implementation formalizes the following concepts:
%
\newcommand{\sourcebasepath}{http://web.student.chalmers.se/~hanghj/cat/doc/html/}
\newcommand{\sourcelink}[1]{\href{\sourcebasepath#1.html}{\texttt{#1}}}
\begin{center}
\begin{tabular}{ l l }
Name & Link \\
Name & Module \\
\hline
Equivalences & \sourcelink{Cat.Equivalence} \\
Categories & \sourcelink{Cat.Category} \\
@ -30,7 +45,7 @@ Furthermore the following items have been partly formalized:
%
\begin{center}
\begin{tabular}{ l l }
Name & Link \\
Name & Module \\
\hline
Category of categories & \sourcelink{Cat.Categories.Cat} \\
Category of relations & \sourcelink{Cat.Categories.Rel} \\
@ -40,7 +55,7 @@ Monoids & \sourcelink{Cat.Category.Monoid} \\
\end{tabular}
\end{center}
%
As well as a range of various results about these. E.g. I have shown
As well as a range of various results about these. E.g.\ I have shown
that the category of sets has products. In the following I aim to
demonstrate some of the techniques employed in this formalization and
in the interest of brevity I will not detail all the things I have
@ -64,15 +79,10 @@ where one can reason about two categories by simply focusing on the data. This
is achieved by creating a function embodying the ``equality principle'' for a
given type.
For the rest of this chapter I will present some of these results. For didactic
reasons no source-code has been included in this chapter. To see the formal
definitions excerpts of the implementation have been included in appendix
\S\ref{ch:app-sources}:
Appendix
\S\ref{sec:app-categories} corresponds to section \S\ref{sec:categories},
appendix \S\ref{sec:app-products} to section \S\ref{sec:products}
and appendix \S\ref{sec:app-monads} to section \S\ref{sec:monads}.
For the rest of this chapter I will present some of these results. For
didactic reasons no source-code has been included in this chapter. To
see the formal definitions the reader is referred to the
implementation which is linked in \S\ref{ch:implementation}.
\section{Categories}
\label{sec:categories}
@ -191,7 +201,7 @@ So the proof goes like this: We `eliminate' the 3 function abstractions by
applying $\propPi$ three times. So our proof obligation becomes:
%
$$
\isProp \left( \id \comp f \equiv f \x f \comp \id \equiv f \right)
\isProp\ \left( \id \comp f \equiv f \x f \comp \id \equiv f \right)
$$
%
Then we eliminate the (non-dependent) sigma-type by applying $\propSig$ giving
@ -224,19 +234,21 @@ Each corresponding to the first three laws for categories. Note that since
$\IsPreCategory$ is not formulated with a chain of sigma-types we wont have any
combinators available to help us here. In stead the paths must be used directly.
\ref{eq:propIsPreCategory} is judgmentally the same as
\ref{eq:propIsPreCategory} is judgmentally the same as:
%
$$
\prod_{a\ b \tp \IsPreCategory} a \equiv b
$$
%
So let $a\ b \tp \IsPreCategory$ be given. To prove the equality $a \equiv b$ is
to give a continuous path from the index-type into the path-space. I.e. a
function $I \to \IsPreCategory$. This path must satisfy being being judgmentally
the same as $a$ at the left endpoint and $b$ at the right endpoint. We know we
can form a continuous path between all projections of $a$ and $b$, this follows
from the type of all the projections being mere propositions. For instance, the
path between $a.\isIdentity$ and $b.\isIdentity$ is simply formed by:
So to prove the proposition let $a\ b \tp \IsPreCategory$ be given. To
prove the equality $a \equiv b$ is to give a continuous path from the
index-type into the path-space. I.e.\ a function $I \to
\IsPreCategory$. This path must satisfy being being judgmentally the
same as $a$ at the left endpoint and $b$ at the right endpoint. We
know we can form a continuous path between all projections of $a$ and
$b$, this follows from the type of all the projections being mere
propositions. For instance, the path between $a.\isIdentity$ and
$b.\isIdentity$ is simply formed by:
%
$$
\propIsIdentity\ a.\isIdentity\ b.\isIdentity
@ -270,18 +282,19 @@ theory, namely that you have to wrap and unwrap equalities at different levels.
It is worth noting that proving this theorem with the regular inductive equality
type would already not be possible, since we at least need extensionality (the
projections are all $\prod$-types). Assuming we had functional extensionality
available to us as an axiom, we would use functional extensionality (in
reverse?) to retrieve the equalities in $a$ and $b$, pattern-match on them to
available to us as an axiom, we would use functional extensionality \TODO{in
reverse?} to retrieve the equalities in $a$ and $b$, pattern-match on them to
see that they are both $\refl$ and then close the proof with $\refl$.
Of course this theorem is not so interesting in the setting of ITT since we know
a priori that equality proofs are unique.
The situation is a bit more complicated when we have a dependent type. For
instance, when we want to show that $\IsCategory$ is a mere proposition.
$\IsCategory$ is a record with two fields, a witness to being a pre category and
the univalence condition. Recall that the univalence condition is indexed by the
identity-proof. So to follow the same recipe as above, let $a\ b \tp
\IsCategory$ be given, to show them equal, we now need to give two paths. One homogeneous:
The situation is a bit more complicated when we have a dependent type.
For instance, when we want to show that $\IsCategory$ is a mere
proposition. The type $\IsCategory$ is a record with two fields, a
witness to being a pre category and the univalence condition. Recall
that the univalence condition is indexed by the identity-proof. So to
follow the same recipe as above, let $a\ b \tp \IsCategory$ be given,
to show them equal, we now need to give two paths. One homogeneous:
%
$$
p \tp a.\isPreCategory \equiv b.\isPreCategory
@ -340,7 +353,7 @@ The usual notion of a function $f \tp A \to B$ having an inverses is:
%
\begin{equation}
\label{eq:isomorphism}
\sum_{g \tp B \to A} f \comp g \equiv \identity_{B} \x g \comp f \equiv \identity_{A}
\sum_{g \tp B \to A} f \comp g \equiv \identity \x g \comp f \equiv \identity
\end{equation}
%
This is defined in \cite[p. 129]{hott-2013} where it is referred to as the a
@ -384,9 +397,10 @@ I give its definition here mainly for completeness, because as I stated we can
move away from this specific instantiation and think about it more abstractly
once we have shown that this definition actually works as an equivalence.
$\var{fromIso}$ can be found in \cite{cubical-agda} where it is known as
$\var{gradLemma}$. The implementation of $\var{fromIso}$ as well as the proof
that this equivalence is a proposition (\ref{eq:propIsEquiv}) can be found in my
The implementation of $\var{fromIso}$ can be found in
\cite{cubical-agda} where it is known as $\var{gradLemma}$. The
implementation of $\var{fromIso}$ as well as the proof that this
equivalence is a proposition (\ref{eq:propIsEquiv}) can be found in my
implementation.
We say that two types $A\;B \tp \Type$ are equivalent exactly if there exists an
@ -586,27 +600,27 @@ go between $\wideoverbar{\approxeq}$ and $\approxeq$.
An inhabitant of $A \approxeq B$ is simply an arrow $f \tp \Arrow\ A\ B$
and its inverse $g \tp \Arrow\ B\ A$. In the opposite category $g$ will
play the role of the isomorphism and $f$ will be the inverse. Similarly we can
go in the opposite direction. I name these maps $\shuffle \tp (A \approxeq
go in the opposite direction. I name these maps $\shufflef \tp (A \approxeq
B) \to (A \wideoverbar{\approxeq} B)$ and $\shuffle^{-1} \tp (A
\wideoverbar{\approxeq} B) \to (A \approxeq B)$ respectively.
As the inverse of $\wideoverbar{\idToIso}$ I will pick $\wideoverbar{\isoToId}
\defeq \isoToId \comp \shuffle$. The proof that they are inverses go as
\defeq \isoToId \comp \shufflef$. The proof that they are inverses go as
follows:
%
\begin{align*}
\wideoverbar{\isoToId} \comp \wideoverbar{\idToIso} & =
\isoToId \comp \shuffle \comp \wideoverbar{\idToIso}
\isoToId \comp \shufflef \comp \wideoverbar{\idToIso}
\\
%% ≡⟨ cong (λ φ → φ x) (cong (λ φ → η ⊙ shuffle ⊙ φ) (funExt lem)) ⟩ \\
%
& \equiv
\isoToId \comp \shuffle \comp \inv{\shuffle} \comp \idToIso
\isoToId \comp \shufflef \comp \inv{\shufflef} \comp \idToIso
&& \text{lemma} \\
%% ≡⟨⟩ \\
& \equiv
\isoToId \comp \idToIso
&& \text{$\shuffle$ is an isomorphism} \\
&& \text{$\shufflef$ is an isomorphism} \\
& \equiv
\identity
&& \text{$\isoToId$ is an isomorphism}
@ -615,7 +629,7 @@ follows:
The other direction is analogous.
The lemma used in step 2 of this proof states that $\wideoverbar{idToIso} \equiv
\inv{\shuffle} \comp \idToIso$. This is a rather straight-forward proof
\inv{\shufflef} \comp \idToIso$. This is a rather straight-forward proof
since being-an-inverse-of is a proposition, so it suffices to show that their
first components are equal, but this holds judgmentally.
@ -681,7 +695,7 @@ together. Step one will be proven with the following lemma:
%
\begin{align}
\label{eq:equivPropSig}
\left(\prod_{a \tp A} \isProp (P\ a)\right) \to \prod_{x\;y \tp \sum_{a \tp A} P\ a} (x \equiv y) \simeq (\fst\ x \equiv \fst\ y)
\left(\prod_{a \tp A} \isProp\ (P\ a)\right) \to \prod_{x\;y \tp \sum_{a \tp A} P\ a} (x \equiv y) \simeq (\fst\ x \equiv \fst\ y)
\end{align}
%
The lemma states that for pairs whose second component are mere propositions
@ -705,7 +719,7 @@ also be equivalent. For our purposes $P \defeq \isEquiv\ A\ B$ and $Q \defeq
\prod_{f \tp A \to B} \left( \isEquiv\ A\ B\ f \simeq \Isomorphism\ f \right)
\end{align}
First, lets prove \ref{eq:equivPropSig}: Let $propP \tp \prod_{a \tp A} \isProp (P\ a)$ and $x\;y \tp \sum_{a \tp A} P\ a$ be given. Because
First, lets prove \ref{eq:equivPropSig}: Let $propP \tp \prod_{a \tp A} \isProp\ (P\ a)$ and $x\;y \tp \sum_{a \tp A} P\ a$ be given. Because
of $\var{fromIsomorphism}$ it suffices to give an isomorphism between
$x \equiv y$ and $\fst\ x \equiv \fst\ y$:
%
@ -770,7 +784,7 @@ end, let $X\;Y \tp \Isomorphism\ f$ be given. Name the maps $x\;y \tp B
paths: $p \tp x \equiv y$ and $\Path\ (\lambda\; i \to
\var{AreInverses}\ f\ (p\ i))\ \mathcal{X}\ \mathcal{Y}$ where $\mathcal{X}$
and $\mathcal{Y}$ denotes the witnesses that $x$ (respectively $y$) is an
inverse to $f$. $p$ is inhabited by:
inverse to $f$. The path $p$ is inhabited by:
%
\begin{align*}
x
@ -844,9 +858,6 @@ that there exists a unique arrow $\pi \tp \Arrow\ X\ (A \x B)$ satisfying
$\pi$ is called the product (arrow) of $f$ and $g$.
\subsection{Span category}
\newcommand\pairA{\mathcal{A}}
\newcommand\pairB{\mathcal{B}}
Given a base category $\bC$ and two objects in this category $\pairA$ and
$\pairB$ we can construct the \nomenindex{span category}:
@ -1018,7 +1029,7 @@ isomorphism, and create a path from this:
\begin{align}
\label{eq:univ-2}
\begin{split}
\var{iso} \tp & X \cong Y \\
\var{iso} \tp & X \approxeq Y \\
& \Path\ (λ i → \Arrow\ (\widetilde{p}\ i)\ \mathcal{A})\ x_{\mathcal{A}}\ y_{\mathcal{A}} \\
& \Path\ (λ i → \Arrow\ (\widetilde{p}\ i)\ \mathcal{B})\ x_{\mathcal{B}}\ y_{\mathcal{B}}
\end{split}
@ -1084,7 +1095,7 @@ To show that this choice fits the bill I must now verify that it satisfies
y_{\mathcal{A}} \lll f ≡ x_{\mathcal{A}} × y_{\mathcal{B}} \lll f ≡ x_{\mathcal{B}}
\end{align}
%
Which, since $f$ is an isomorphism and $p_{\mathcal{A}}$ (resp. $p_{\mathcal{B}}$)
Which, since $f$ is an isomorphism and $p_{\mathcal{A}}$ (resp.\ $p_{\mathcal{B}}$)
is a path varying according to a path constructed from this isomorphism, this is
exactly what \ref{eq:domain-twist-0} gives us.
%
@ -1100,7 +1111,7 @@ This concludes the first direction of the isomorphism that we are constructing.
For the other direction we are given the isomorphism:
%
$$
(f, \inv{f}, \var{inv}_f)
(f, \inv{f}, \var{inv}_f, \var{inv}_{\inv{f}})
\tp
(X, x_{\mathcal{A}}, x_{\mathcal{B}}) \approxeq (Y, y_{\mathcal{A}}, y_{\mathcal{B}})
$$
@ -1108,7 +1119,10 @@ $$
Projecting out the first component gives us the isomorphism
%
$$
(\fst\ f, \fst\ \inv{f}, \congruence\ \fst\ \var{inv}_f, \congruence\ \fst\ \var{inv}_{\inv{f}})
(\fst\ f, \fst\ \inv{f}
, \congruence\ \fst\ \var{inv}_f
, \congruence\ \fst\ \var{inv}_{\inv{f}}
)
\tp X \approxeq Y
$$
%
@ -1141,11 +1155,11 @@ This is achieved with the following lemma:
%
Which is used without proof. See the implementation for the details.
\ref{eq:product-paths} is the proven with the propositions:
\ref{eq:product-paths} is then proven with the propositions:
%
\begin{align}
\begin{split}
\label{eq:product-paths}
%% \label{eq:product-paths}
\var{coe}\ \widetilde{p}_{\mathcal{A}}\ x_{\mathcal{A}} ≡ y_{\mathcal{A}}\\
\var{coe}\ \widetilde{p}_{\mathcal{B}}\ x_{\mathcal{B}} ≡ y_{\mathcal{B}}
\end{split}
@ -1155,7 +1169,7 @@ The proof of the first one is:
%
\begin{align*}
\var{coe}\ \widetilde{p}_{\mathcal{A}}\ x_{\mathcal{A}}
& ≡ x_{\mathcal{A}} \lll \fst\ \inv{f} && \text{$\var{coeDom}$ and the isomorphism $f, \inv{f}$} \\
& ≡ x_{\mathcal{A}} \lll \fst\ \inv{f} && \text{\ref{eq:coeDom} and the isomorphism $f, \inv{f}$} \\
& ≡ y_{\mathcal{A}} && \text{\ref{eq:pairArrowLaw} for $\inv{f}$}
\end{align*}
%
@ -1167,10 +1181,10 @@ gory details.
%
\subsection{Propositionality of products}
%
Now that we have constructed the span category\index{span category} I will
demonstrate how to use this to prove that products are propositional. I will
do this by showing that terminal objects in this category are equivalent to
products:
Now that we have constructed the span category\index{span category} I
will demonstrate how to use this to prove that products are
propositional. I will do this by showing that terminal objects in this
category are equivalent to products:
%
\begin{align}
\var{Terminal} ≃ \var{Product}\ ℂ\ \mathcal{A}\ \mathcal{B}
@ -1179,7 +1193,7 @@ Now that we have constructed the span category\index{span category} I will
And as always we do this by constructing an isomorphism:
%
In the direction $\var{Terminal} → \var{Product}\ ℂ\ \mathcal{A}\ \mathcal{B}$
we are given a terminal object $X, x_𝒜, x_ℬ$. $X$ Will be the product-object and
we are given a terminal object $X, x_𝒜, x_ℬ$. $X$ will be the product-object and
$x_𝒜, x_ℬ$ will be the product arrows, so it just remains to verify that this is
indeed a product. That is, for an object $Y$ and two arrows $y_𝒜 \tp
\Arrow\ Y\ 𝒜$, $y_ℬ\ \Arrow\ Y\ ℬ$ we must find a unique arrow $f \tp
@ -1265,8 +1279,8 @@ The monoidal formulation of monads consists of the following data:
\label{eq:monad-monoidal-data}
\begin{split}
\EndoR & \tp \Endo ℂ \\
\pure & \tp \NT{\EndoR^0}{\EndoR} \\
\join & \tp \NT{\EndoR^2}{\EndoR}
\pureNT & \tp \NT{\EndoR^0}{\EndoR} \\
\joinNT & \tp \NT{\EndoR^2}{\EndoR}
\end{split}
\end{align}
%
@ -1299,11 +1313,11 @@ The Kleisli-formulation consists of the following data:
\begin{align}
\begin{split}
\label{eq:monad-kleisli-data}
\EndoR & \tp \Object → \Object \\
\omapR & \tp \Object → \Object \\
\pure & \tp % \prod_{X \tp Object}
\Arrow\ X\ (\EndoR\ X) \\
\bind & \tp % \prod_{X\;Y \tp Object} → \Arrow\ X\ (\EndoR\ Y)
\Arrow\ (\EndoR\ X)\ (\EndoR\ Y)
\Arrow\ X\ (\omapR\ X) \\
\bind & \tp % \prod_{X\;Y \tp Object} → \Arrow\ X\ (\omapR\ Y)
\Arrow\ (\omapR\ X)\ (\omapR\ Y)
\end{split}
\end{align}
%
@ -1346,22 +1360,20 @@ In the monoidal formulation we can define $\bind$:
\newcommand\pureX{\wideoverbar{\pure}}%
\newcommand\fmapX{\wideoverbar{\fmap}}%
\begin{align}
\bind\ f \defeq \joinX \lll \fmap\ f
\bind\ f \defeq \join \lll \fmap\ f
\end{align}
%
And likewise in the Kleisli formulation we can define $\join$:
%
\begin{align}
\join \defeq \bindX\ \identity
\join \defeq \bind\ \identity
\end{align}
%
Here $\joinX$ corresponds to the arrow from the natural
transformation $\join$. $\bindX$ on the other hand corresponds to a
natural transformation constructed from $\bind$. It now remains to show that
this construction indeed gives rise to a monad. This will be done in two steps.
First we will assume that we have a monad in the monoidal form; $(\EndoR, \pure,
\join)$ and then show that $\EndoR, \pure, \bind$ is indeed a monad in the
Kleisli form. In the second part we will show the other direction.
It now remains to show that this construction indeed gives rise to a
monad. This will be done in two steps. First we will assume that we
have a monad in the monoidal form; $(\EndoR, \pure, \join)$ and then
show that $(\omapR, \pure, \bind)$ is indeed a monad in the Kleisli
form. In the second part we will show the other direction.
\subsubsection{Monoidal to Kleisli}
Let $(\EndoR, \pure, \join)$ be given as in \ref{eq:monad-monoidal-data}
@ -1376,11 +1388,11 @@ formulation we pick:
\end{split}
\end{align}
%
$\EndoRX$ is the object map of the endo-functor $\EndoR$,
$\pureX$ and $\joinX$ are the arrows from the natural
transformations $\pure$ and $\join$ respectively. $\fmapX$ is the
arrow map of the endo-functor $\EndoR$. It now just remains to verify
the laws \kleislilaws. For \ref{eq:monad-kleisli-laws-0}:
$\EndoRX$ is the object map of the endo-functor $\EndoR$, $\pureX$ and
$\joinX$ are the arrows from the natural transformations $\pure$ and
$\join$ respectively. The term $\fmapX$ is the arrow map of the
endo-functor $\EndoR$. It now just remains to verify the laws
\kleislilaws. For \ref{eq:monad-kleisli-laws-0}:
%
\begin{align*}
\bind\ \pure & ≡

27
doc/macros.tex
View File

@ -72,13 +72,17 @@
\newcommand\isequiv{\varindex{isequiv}}
\newcommand\qinv{\varindex{qinv}}
\newcommand\fiber{\varindex{fiber}}
\newcommand\shuffle{\varindex{shuffle}}
\newcommand\shufflef{\varindex{shuffle}}
\newcommand\Univalent{\varindex{Univalent}}
\newcommand\refl{\varindex{refl}}
\newcommand\isoToId{\varindex{isoToId}}
\newcommand\Isomorphism{\varindex{Isomorphism}}
\newcommand\rrr{\ggg}
\newcommand\fish{\mathrel{\wideoverbar{\rrr}}}
%% \newcommand\fish{\mathbin{↣}}
%% \newcommand\fish{\mathbin{⤅}}
\newcommand\fish{\mathbin{⤇}}
%% \newcommand\fish{\mathbin{⤜}}
%% \newcommand\fish{\mathrel{\wideoverbar{\rrr}}}
\newcommand\fst{\varindex{fst}}
\newcommand\snd{\varindex{snd}}
\newcommand\Path{\varindex{Path}}
@ -87,12 +91,29 @@
\newcommand*{\QED}{\hfill\ensuremath{\square}}%
\newcommand\uexists{\exists!}
\newcommand\Arrow{\varindex{Arrow}}
\newcommand\embellish[1]{\widehat{#1}}
\newcommand\nattrans[1]{\embellish{#1}}
\newcommand\functor[1]{\embellish{#1}}
\newcommand\NTsym{\varindex{NT}}
\newcommand\NT[2]{\NTsym\ #1\ #2}
\newcommand\Endo[1]{\varindex{Endo}\ #1}
\newcommand\EndoR{\mathcal{R}}
\newcommand\EndoR{\functor{\mathcal{R}}}
\newcommand\omapR{\mathcal{R}}
\newcommand\funExt{\varindex{funExt}}
\newcommand{\suc}[1]{\varindex{suc}\ #1}
\newcommand{\trans}{\varindex{trans}}
\newcommand{\toKleisli}{\varindex{toKleisli}}
\newcommand{\toMonoidal}{\varindex{toMonoidal}}
\newcommand\pairA{\mathcal{A}}
\newcommand\pairB{\mathcal{B}}
\newcommand{\joinNT}{\functor{\varindex{join}}}
\newcommand{\pureNT}{\functor{\varindex{pure}}}
\newcommand{\hrefsymb}[2]{\href{#1}{#2 \ExternalLink}}
\newcommand{\sourcebasepath}{http://web.student.chalmers.se/\textasciitilde hanghj/cat/doc/html/}
\newcommand{\docbasepath}{https://github.com/fredefox/cat/}
\newcommand{\sourcelink}[1]{\hrefsymb
{\sourcebasepath#1.html}
{\texttt{#1}}
}
\newcommand{\gitlink}{\hrefsymb{\docbasepath}{\texttt{\docbasepath}}}
\newcommand{\doclink}{\hrefsymb{\sourcebasepath}{\texttt{\sourcebasepath}}}

10
doc/main.tex
View File

@ -62,12 +62,12 @@
\nocite{coquand-2013}
\bibliography{refs}
\begin{appendices}
\setcounter{page}{1}
\pagenumbering{roman}
\input{sources.tex}
%% \begin{appendices}
%% \setcounter{page}{1}
%% \pagenumbering{roman}
%% \input{sources.tex}
%% \input{planning.tex}
%% \input{halftime.tex}
\end{appendices}
%% \end{appendices}
\printindex
\end{document}

28
doc/packages.tex
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@ -4,14 +4,17 @@
\bibliographystyle{plain}
\usepackage{xcolor}
%% \mode<report>{
\usepackage[
%% hidelinks,
pdfusetitle,
pdfsubject={category theory},
pdfkeywords={type theory, homotopy theory, category theory, agda}]
{hyperref}
\definecolor{darkorange}{HTML}{ff8c00}
\hypersetup{allbordercolors={darkorange}}
%% }
%% \definecolor{darkorange}{HTML}{ff8c00}
%% \hypersetup{allbordercolors={darkorange}}
\hypersetup{hidelinks}
\usepackage{graphicx}
\usepackage{parskip}
@ -116,3 +119,24 @@
\makeatother
\usepackage{xspace}
\usepackage{tikz}
\newcommand{\ExternalLink}{%
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@ -1,41 +1,49 @@
\documentclass[a4paper,handout]{beamer}
\beamertemplatenavigationsymbolsempty
%% \usecolortheme[named=seagull]{structure}
\input{packages.tex}
\input{macros.tex}
\title{Univalent Categories}
\author{Frederik Hangh{\o}j Iversen}
\title[Univalent Categories]{Univalent Categories\\ \footnotesize A formalization of category theory in Cubical Agda}
\newcommand{\myname}{Frederik Hangh{\o}j Iversen}
\author[\myname]{
\myname\\
\footnotesize Supervisors: Thierry Coquand, Andrea Vezzosi\\
Examiner: Andreas Abel
}
\institute{Chalmers University of Technology}
\begin{document}
\frame{\titlepage}
\begin{frame}
\frametitle{Motivating example}
\framesubtitle{Functional extensionality}
Consider the functions
\begin{align*}
\var{zeroLeft} & \defeq (n \tp \bN) \mto (0 + n \tp \bN) \\
\var{zeroRight} & \defeq (n \tp \bN) \mto (n + 0 \tp \bN)
\end{align*}
\pause
We have
%
$$
\prod_{n \tp \bN} n + 0 \equiv 0 + n
$$
%
\pause
But not
%
$$
\var{zeroLeft} \equiv \var{zeroRight}
$$
%
\pause
We need
%
$$
\funExt \tp \prod_{a \tp A} f\ a \equiv g\ a \to f \equiv g
$$
Consider the functions
\begin{align*}
\var{zeroLeft} & \defeq \lambda (n \tp \bN) \mto (0 + n \tp \bN) \\
\var{zeroRight} & \defeq \lambda (n \tp \bN) \mto (n + 0 \tp \bN)
\end{align*}
\pause
We have
%
$$
\prod_{n \tp \bN} \var{zeroLeft}\ n \equiv \var{zeroRight}\ n
$$
%
\pause
But not
%
$$
\var{zeroLeft} \equiv \var{zeroRight}
$$
%
\pause
We need
%
$$
\funExt \tp \prod_{a \tp A} f\ a \equiv g\ a \to f \equiv g
$$
\end{frame}
\begin{frame}
\frametitle{Motivating example}
@ -52,8 +60,8 @@ $$
We need univalence:
$$(A \simeq B) \simeq (A \equiv B)$$
\pause
%
We will return to $\simeq$, but for not, think of it as an
%
We will return to $\simeq$, but for now think of it as an
isomorphism, so it induces maps:
\begin{align*}
\var{toPath} & \tp (A \simeq B) \to (A \equiv B) \\
@ -63,174 +71,579 @@ $$
\begin{frame}
\frametitle{Paths}
\framesubtitle{Definition}
Heterogeneous paths
\begin{equation*}
\Path \tp (P \tp I → \MCU) → P\ 0 → P\ 1 → \MCU
\end{equation*}
\pause
Heterogeneous paths
\begin{equation*}
\Path \tp (P \tp I → \MCU) → P\ 0 → P\ 1 → \MCU
\end{equation*}
\pause
For $P \tp I \to \MCU$, $A \tp \MCU$ and $a_0, a_1 \tp A$
inhabitants of $\Path\ P\ a_0\ a_1$ are like functions
%
$$
p \tp \prod_{i \tp I} P\ i
$$
%
Which satisfy $p\ 0 & = a_0$ and $p\ 1 & = a_1$
\pause
%
$$
p \tp \prod_{i \tp I} P\ i
$$
%
Which satisfy $p\ 0 & = a_0$ and $p\ 1 & = a_1$
\pause
Homogenous paths
$$
a_0 \equiv a_1 \defeq \Path\ (\var{const}\ A)\ a_0\ a_1
$$
Homogenous paths
$$
a_0 \equiv a_1 \defeq \Path\ (\var{const}\ A)\ a_0\ a_1
$$
\end{frame}
\begin{frame}
\frametitle{Paths}
\framesubtitle{Functional extenstionality}
$$
\funExt & \tp \prod_{a \tp A} f\ a \equiv g\ a \to f \equiv g
$$
\pause
$$
\funExt\ p \defeq λ i\ a → p\ a\ i
$$
\pause
$$
\funExt\ (\var{const}\ \refl)
\tp
\var{zeroLeft} \equiv \var{zeroRight}
$$
\frametitle{Paths}
\framesubtitle{Functional extenstionality}
$$
\funExt & \tp \prod_{a \tp A} f\ a \equiv g\ a \to f \equiv g
$$
\pause
$$
\funExt\ p \defeq λ i\ a → p\ a\ i
$$
\pause
$$
\funExt\ (\var{const}\ \refl)
\tp
\var{zeroLeft} \equiv \var{zeroRight}
$$
\end{frame}
\begin{frame}
\frametitle{Paths}
\framesubtitle{Homotopy levels}
\begin{align*}
& \isContr && \tp \MCU \to \MCU \\
& \isContr\ A && \defeq \sum_{c \tp A} \prod_{a \tp A} a \equiv c
\end{align*}
\pause
\begin{align*}
& \isProp && \tp \MCU \to \MCU \\
& \isProp\ A && \defeq \prod_{a_0, a_1 \tp A} a_0 \equiv a_1
\end{align*}
\pause
\begin{align*}
& \isSet && \tp \MCU \to \MCU \\
& \isSet\ A && \defeq \prod_{a_0, a_1 \tp A} \isProp\ (a_0 \equiv a_1)
\end{align*}
\pause
\begin{align*}
& \isContr && \tp \MCU \to \MCU \\
& \isContr\ A && \defeq \sum_{c \tp A} \prod_{a \tp A} a \equiv c
\end{align*}
\pause
\begin{align*}
& \isProp && \tp \MCU \to \MCU \\
& \isProp\ A && \defeq \prod_{a_0, a_1 \tp A} a_0 \equiv a_1
\end{align*}
\pause
\begin{align*}
& \isSet && \tp \MCU \to \MCU \\
& \isSet\ A && \defeq \prod_{a_0, a_1 \tp A} \isProp\ (a_0 \equiv a_1)
\end{align*}
\begin{align*}
& \isGroupoid && \tp \MCU \to \MCU \\
& \isGroupoid\ A && \defeq \prod_{a_0, a_1 \tp A} \isSet\ (a_0 \equiv a_1)
\end{align*}
\end{frame}
\begin{frame}
\frametitle{Paths}
\framesubtitle{A few lemmas}
Let $D$ be a type-family:
$$
D \tp \prod_{b \tp A} \prod_{p \tp a ≡ b} \MCU
$$
%
\pause
And $d$ and in inhabitant of $D$ at $\refl$:
%
$$
d \tp D\ a\ \refl
$$
%
\pause
We then have the function:
%
\begin{equation}
\pathJ\ D\ d \tp \prod_{b \tp A} \prod_{p \tp a ≡ b} D\ a\ p
\end{equation}
\frametitle{Paths}
\framesubtitle{A few lemmas}
Let $D$ be a type-family:
$$
D \tp \prod_{b \tp A} \prod_{p \tp a ≡ b} \MCU
$$
%
\pause
And $d$ and in inhabitant of $D$ at $\refl$:
%
$$
d \tp D\ a\ \refl
$$
%
\pause
We then have the function:
%
$$
\pathJ\ D\ d \tp \prod_{b \tp A} \prod_{p \tp a ≡ b} D\ b\ p
$$
\end{frame}
\begin{frame}
\frametitle{Paths}
\framesubtitle{A few lemmas}
Given
\begin{align*}
A & \tp \MCU \\
P & \tp A \to \MCU \\
\var{propP} & \tp \prod_{x \tp A} \isProp\ (P\ x) \\
p & \tp a_0 \equiv a_1 \\
p_0 & \tp P\ a_0 \\
p_1 & \tp P\ a_1
\end{align*}
%
We have
$$
\lemPropF\ \var{propP}\ p
\tp
\Path\ (\lambda\; i \mto P\ (p\ i))\ p_0\ p_1
$$
%
\frametitle{Paths}
\framesubtitle{A few lemmas}
Given
\begin{align*}
A & \tp \MCU \\
P & \tp A \to \MCU \\
\var{propP} & \tp \prod_{x \tp A} \isProp\ (P\ x) \\
p & \tp a_0 \equiv a_1 \\
p_0 & \tp P\ a_0 \\
p_1 & \tp P\ a_1
\end{align*}
%
We have
$$
\lemPropF\ \var{propP}\ p
\tp
\Path\ (\lambda\; i \mto P\ (p\ i))\ p_0\ p_1
$$
%
\end{frame}
\begin{frame}
\frametitle{Paths}
\framesubtitle{A few lemmas}
$\prod$ preserves $\isProp$:
$$
\mathit{propPi}
\tp
\left(\prod_{a \tp A} \isProp\ (P\ a)\right)
\to \isProp\ \left(\prod_{a \tp A} P\ a\right)
$$
\pause
$\sum$ preserves $\isProp$:
$$
\mathit{propSig} \tp \isProp\ A \to \left(\prod_{a \tp A} \isProp\ (P\ a)\right) \to \isProp\ \left(\sum_{a \tp A} P\ a\right)
$$
\frametitle{Paths}
\framesubtitle{A few lemmas}
$\prod$ preserves $\isProp$:
$$
\mathit{propPi}
\tp
\left(\prod_{a \tp A} \isProp\ (P\ a)\right)
\to \isProp\ \left(\prod_{a \tp A} P\ a\right)
$$
\pause
$\sum$ preserves $\isProp$:
$$
\mathit{propSig} \tp \isProp\ A \to \left(\prod_{a \tp A} \isProp\ (P\ a)\right) \to \isProp\ \left(\sum_{a \tp A} P\ a\right)
$$
\end{frame}
\begin{frame}
\frametitle{Categories}
\framesubtitle{Definition}
Data:
\begin{align*}
\Object & \tp \Type \\
\Arrow & \tp \Object \to \Object \to \Type \\
\identity & \tp \Arrow\ A\ A \\
\lll & \tp \Arrow\ B\ C \to \Arrow\ A\ B \to \Arrow\ A\ C
\end{align*}
%
Laws:
%
$$
h \lll (g \lll f) ≡ (h \lll g) \lll f
$$
$$
\identity \lll f ≡ f \x
f \lll \identity ≡ f
$$
\pause
1-categories:
$$
\isSet\ (\Arrow\ A\ B)
$$
\pause
Univalent categories:
$$
\isEquiv\ (A \equiv B)\ (A \approxeq B)\ \idToIso
$$
\frametitle{Pre categories}
\framesubtitle{Definition}
Data:
\begin{align*}
\Object & \tp \Type \\
\Arrow & \tp \Object \to \Object \to \Type \\
\identity & \tp \Arrow\ A\ A \\
\lll & \tp \Arrow\ B\ C \to \Arrow\ A\ B \to \Arrow\ A\ C
\end{align*}
%
\pause
Laws:
%
$$
h \lll (g \lll f) ≡ (h \lll g) \lll f
$$
$$
(\identity \lll f ≡ f)
\x
(f \lll \identity ≡ f)
$$
\pause
1-categories:
$$
\isSet\ (\Arrow\ A\ B)
$$
\end{frame}
\begin{frame}
\frametitle{Categories}
\framesubtitle{Univalence}
\begin{align*}
\var{IsIdentity} & \defeq
\prod_{A\ B \tp \Object} \prod_{f \tp \Arrow\ A\ B} \phi\ f
%% \\
%% & \mathrel{\ } \identity \lll f \equiv f \x f \lll \identity \equiv f
\end{align*}
where
$$
\phi\ f \defeq \identity \lll f \equiv f \x f \lll \identity \equiv f
$$
Let $\approxeq$ denote ismorphism of objects. We can then construct
the identity isomorphism in any category:
$$
\identity , \identity , \var{isIdentity} \tp A \approxeq A
$$
Likewise since paths are substitutive we can promote a path to an isomorphism:
$$
\idToIso \tp A ≡ B → A ≊ B
$$
For a category to be univalent we require this to be an equivalence:
\frametitle{Pre categories}
\framesubtitle{Propositionality}
$$
\isProp\ \left( (\identity \comp f \equiv f) \x (f \comp \identity \equiv f) \right)
$$
\pause
\begin{align*}
\isProp\ \IsPreCategory
\end{align*}
\pause
\begin{align*}
\var{isAssociative} & \tp \var{IsAssociative}\\
\isIdentity & \tp \var{IsIdentity}\\
\var{arrowsAreSets} & \tp \var{ArrowsAreSets}
\end{align*}
\pause
\begin{align*}
& \var{propIsAssociative} && a.\var{isAssociative}\
&& b.\var{isAssociative} && i \\
& \propIsIdentity && a.\isIdentity\
&& b.\isIdentity && i \\
& \var{propArrowsAreSets} && a.\var{arrowsAreSets}\
&& b.\var{arrowsAreSets} && i
\end{align*}
\end{frame}
\begin{frame}
\frametitle{Categories}
\framesubtitle{Univalence}
\begin{align*}
\var{IsIdentity} & \defeq
\prod_{A\ B \tp \Object} \prod_{f \tp \Arrow\ A\ B} \phi\ f
%% \\
%% & \mathrel{\ } \identity \lll f \equiv f \x f \lll \identity \equiv f
\end{align*}
where
$$
\phi\ f \defeq \identity \lll f \equiv f \x f \lll \identity \equiv f
$$
\pause
Let $\approxeq$ denote ismorphism of objects. We can then construct
the identity isomorphism in any category:
$$
\identity , \identity , \var{isIdentity} \tp A \approxeq A
$$
\pause
Likewise since paths are substitutive we can promote a path to an isomorphism:
$$
\idToIso \tp A ≡ B → A ≊ B
$$
\pause
For a category to be univalent we require this to be an equivalence:
%
$$
\isEquiv\ (A \equiv B)\ (A \approxeq B)\ \idToIso
$$
%
\end{frame}
\begin{frame}
\frametitle{Categories}
\framesubtitle{Univalence, cont'd}
$$\isEquiv\ (A \equiv B)\ (A \approxeq B)\ \idToIso$$
\pause%
$$(A \equiv B) \simeq (A \approxeq B)$$
\pause%
$$(A \equiv B) \cong (A \approxeq B)$$
\pause%
Name the above maps:
$$\idToIso \tp A ≡ B → A ≊ B$$
%
$$\isoToId \tp (A \approxeq B) \to (A \equiv B)$$
\end{frame}
\begin{frame}
\frametitle{Categories}
\framesubtitle{Propositionality}
$$
\isProp\ \IsCategory = \prod_{a, b \tp \IsCategory} a \equiv b
$$
\pause
So, for
$$
a\ b \tp \IsCategory
$$
the proof obligation is the pair:
%
\begin{align*}
p & \tp a.\isPreCategory \equiv b.\isPreCategory \\
& \mathrel{\ } \Path\ (\lambda\; i \to (p\ i).Univalent)\ a.\isPreCategory\ b.\isPreCategory
\end{align*}
\end{frame}
\begin{frame}
\frametitle{Categories}
\framesubtitle{Propositionality, cont'd}
First path given by:
$$
p
\defeq
\var{propIsPreCategory}\ a\ b
\tp
a.\isPreCategory \equiv b.\isPreCategory
$$
\pause
Use $\lemPropF$ for the latter.
\pause
%
Univalence is indexed by an identity proof. So $A \defeq
IsIdentity\ identity$ and $B \defeq \var{Univalent}$.
\pause
%
$$
\lemPropF\ \var{propUnivalent}\ p
$$
\end{frame}
\begin{frame}
\frametitle{Categories}
\framesubtitle{A theorem}
%
Let the isomorphism $(ι, \inv{ι}) \tp A \approxeq B$.
%
\pause
%
The isomorphism induces the path
%
$$
p \defeq \idToIso\ (\iota, \inv{\iota}) \tp A \equiv B
$$
%
\pause
and consequently an arrow:
%
$$
p_{\var{dom}} \defeq \congruence\ (λ x → \Arrow\ x\ X)\ p
\tp
\Arrow\ A\ X \equiv \Arrow\ B\ X
$$
%
\pause
The proposition is:
%
\begin{align}
\label{eq:coeDom}
\tag{$\var{coeDom}$}
\prod_{f \tp A \to X}
\var{coe}\ p_{\var{dom}}\ f \equiv f \lll \inv{\iota}
\end{align}
\end{frame}
\begin{frame}
\frametitle{Categories}
\framesubtitle{A theorem, proof}
\begin{align*}
\var{coe}\ p_{\var{dom}}\ f
& \equiv f \lll \inv{(\idToIso\ p)} && \text{By path-induction} \\
& \equiv f \lll \inv{\iota}
&& \text{$\idToIso$ and $\isoToId$ are inverses}\\
\end{align*}
\pause
%
Induction will be based at $A$. Let $\widetilde{B}$ and $\widetilde{p}
\tp A \equiv \widetilde{B}$ be given.
%
\pause
%
Define the family:
%
$$
D\ \widetilde{B}\ \widetilde{p} \defeq
\var{coe}\ \widetilde{p}_{\var{dom}}\ f
\equiv
f \lll \inv{(\idToIso\ \widetilde{p})}
$$
\pause
%
The base-case becomes:
$$
d \tp D\ A\ \refl =
\var{coe}\ \refl_{\var{dom}}\ f \equiv f \lll \inv{(\idToIso\ \refl)}
$$
\end{frame}
\begin{frame}
\frametitle{Categories}
\framesubtitle{A theorem, proof, cont'd}
$$
d \tp
\var{coe}\ \refl_{\var{dom}}\ f \equiv f \lll \inv{(\idToIso\ \refl)}
$$
\pause
\begin{align*}
\var{coe}\ \refl^*\ f
& \equiv f
&& \text{$\refl$ is a neutral element for $\var{coe}$}\\
& \equiv f \lll \identity \\
& \equiv f \lll \var{subst}\ \refl\ \identity
&& \text{$\refl$ is a neutral element for $\var{subst}$}\\
& \equiv f \lll \inv{(\idToIso\ \refl)}
&& \text{By definition of $\idToIso$}\\
\end{align*}
\pause
In conclusion, the theorem is inhabited by:
$$
\label{eq:pathJ-example}
\pathJ\ D\ d\ B\ p
$$
\end{frame}
\begin{frame}
\frametitle{Span category} \framesubtitle{Definition} Given a base
category $\bC$ and two objects in this category $\pairA$ and $\pairB$
we can construct the \nomenindex{span category}:
%
\pause
Objects:
$$
\sum_{X \tp Object} \Arrow\ X\ \pairA × \Arrow\ X\ \pairB
$$
\pause
%
Arrows between objects $A ,\ a_{\pairA} ,\ a_{\pairB}$ and
$B ,\ b_{\pairA} ,\ b_{\pairB}$:
%
$$
\sum_{f \tp \Arrow\ A\ B}
b_{\pairA} \lll f \equiv a_{\pairA} \x
b_{\pairB} \lll f \equiv a_{\pairB}
$$
\end{frame}
\begin{frame}
\frametitle{Span category}
\framesubtitle{Univalence}
\begin{align*}
\label{eq:univ-0}
(X , x_{\mathcal{A}} , x_{\mathcal{B}}) ≡ (Y , y_{\mathcal{A}} , y_{\mathcal{B}})
\end{align*}
\begin{align*}
\label{eq:univ-1}
\begin{split}
p \tp & X \equiv Y \\
& \Path\ (λ i → \Arrow\ (p\ i)\ \mathcal{A})\ x_{\mathcal{A}}\ y_{\mathcal{A}} \\
& \Path\ (λ i → \Arrow\ (p\ i)\ \mathcal{B})\ x_{\mathcal{B}}\ y_{\mathcal{B}}
\end{split}
\end{align*}
\begin{align*}
\begin{split}
\var{iso} \tp & X \approxeq Y \\
& \Path\ (λ i → \Arrow\ (\widetilde{p}\ i)\ \mathcal{A})\ x_{\mathcal{A}}\ y_{\mathcal{A}} \\
& \Path\ (λ i → \Arrow\ (\widetilde{p}\ i)\ \mathcal{B})\ x_{\mathcal{B}}\ y_{\mathcal{B}}
\end{split}
\end{align*}
\begin{align*}
(X , x_{\mathcal{A}} , x_{\mathcal{B}}) ≊ (Y , y_{\mathcal{A}} , y_{\mathcal{B}})
\end{align*}
\end{frame}
\begin{frame}
\frametitle{Span category}
\framesubtitle{Univalence, proof}
%
\begin{align*}
%% (f, \inv{f}, \var{inv}_f, \var{inv}_{\inv{f}})
%% \tp
(X, x_{\mathcal{A}}, x_{\mathcal{B}}) \approxeq (Y, y_{\mathcal{A}}, y_{\mathcal{B}})
\to
\begin{split}
\var{iso} \tp & X \approxeq Y \\
& \Path\ (λ i → \Arrow\ (\widetilde{p}\ i)\ \mathcal{A})\ x_{\mathcal{A}}\ y_{\mathcal{A}} \\
& \Path\ (λ i → \Arrow\ (\widetilde{p}\ i)\ \mathcal{B})\ x_{\mathcal{B}}\ y_{\mathcal{B}}
\end{split}
\end{align*}
\pause
%
Let $(f, \inv{f}, \var{inv}_f, \var{inv}_{\inv{f}})$ be an inhabitant
of the antecedent.\pause
Projecting out the first component gives us the isomorphism
%
$$
(\fst\ f, \fst\ \inv{f}
, \congruence\ \fst\ \var{inv}_f
, \congruence\ \fst\ \var{inv}_{\inv{f}}
)
\tp X \approxeq Y
$$
\pause
%
This gives rise to the following paths:
%
\begin{align*}
\begin{split}
\widetilde{p} & \tp X \equiv Y \\
\widetilde{p}_{\mathcal{A}} & \tp \Arrow\ X\ \mathcal{A} \equiv \Arrow\ Y\ \mathcal{A} \\
\end{split}
\end{align*}
%
\end{frame}
\begin{frame}
\frametitle{Span category}
\framesubtitle{Univalence, proof, cont'd}
It remains to construct:
%
\begin{align*}
\begin{split}
\label{eq:product-paths}
& \Path\ (λ i → \widetilde{p}_{\mathcal{A}}\ i)\ x_{\mathcal{A}}\ y_{\mathcal{A}}
\end{split}
\end{align*}
\pause
%
This is achieved with the following lemma:
%
\begin{align*}
\prod_{q \tp A \equiv B} \var{coe}\ q\ x_{\mathcal{A}} ≡ y_{\mathcal{A}}
→
\Path\ (λ i → q\ i)\ x_{\mathcal{A}}\ y_{\mathcal{A}}
\end{align*}
%
Which is used without proof.\pause
So the construction reduces to:
%
\begin{align*}
\var{coe}\ \widetilde{p}_{\mathcal{A}}\ x_{\mathcal{A}} ≡ y_{\mathcal{A}}
\end{align*}%
\pause%
This is proven with:
%
\begin{align*}
\var{coe}\ \widetilde{p}_{\mathcal{A}}\ x_{\mathcal{A}}
& ≡ x_{\mathcal{A}} \lll \fst\ \inv{f} && \text{\ref{eq:coeDom}} \\
& ≡ y_{\mathcal{A}} && \text{Property of span category}
\end{align*}
\end{frame}
\begin{frame}
\frametitle{Propositionality of products}
We have
%
$$
\isProp\ \var{Terminal}
$$\pause
%
We can show:
\begin{align*}
\var{Terminal} ≃ \var{Product}\ ℂ\ \mathcal{A}\ \mathcal{B}
\end{align*}
\pause
And since equivalences preserve homotopy levels we get:
%
$$
\isProp\ \left(\var{Product}\ \bC\ \mathcal{A}\ \mathcal{B}\right)
$$
\end{frame}
\begin{frame}
\frametitle{Monads}
\framesubtitle{Monoidal form}
%
\begin{align*}
\EndoR & \tp \Endo ℂ \\
\pureNT
& \tp \NT{\EndoR^0}{\EndoR} \\
\joinNT
& \tp \NT{\EndoR^2}{\EndoR}
\end{align*}
\pause
%
Let $\fmap$ be the map on arrows of $\EndoR$. Likewise
$\pure$ and $\join$ are the maps of the natural transformations
$\pureNT$ and $\joinNT$ respectively.
%
\begin{align*}
\join \lll \fmap\ \join
& ≡ \join \lll \join \\
\join \lll \pure\ & ≡ \identity \\
\join \lll \fmap\ \pure & ≡ \identity
\end{align*}
\end{frame}
\begin{frame}
\frametitle{Monads}
\framesubtitle{Kleisli form}
%
\begin{align*}
\omapR & \tp \Object → \Object \\
\pure & \tp % \prod_{X \tp Object}
\Arrow\ X\ (\omapR\ X) \\
\bind & \tp
\Arrow\ X\ (\omapR\ Y)
\to
\Arrow\ (\omapR\ X)\ (\omapR\ Y)
\end{align*}\pause
%
\begin{align*}
\fish & \tp
\Arrow\ A\ (\omapR\ B)
→
\Arrow\ B\ (\omapR\ C)
→
\Arrow\ A\ (\omapR\ C) \\
f \fish g & \defeq f \rrr (\bind\ g)
\end{align*}
\pause
%
\begin{align*}
\label{eq:monad-kleisli-laws-0}
\bind\ \pure & ≡ \identity_{\omapR\ X} \\
\label{eq:monad-kleisli-laws-1}
\pure \fish f & ≡ f \\
\label{eq:monad-kleisli-laws-2}
(\bind\ f) \rrr (\bind\ g) & ≡ \bind\ (f \fish g)
\end{align*}
\end{frame}
\begin{frame}
\frametitle{Monads}
\framesubtitle{Equivalence}
In the monoidal formulation we can define $\bind$:
%
$$
\bind\ f \defeq \join \lll \fmap\ f
$$
\pause
%
And likewise in the Kleisli formulation we can define $\join$:
%
$$
\join \defeq \bind\ \identity
$$
\pause
The laws are logically equivalent. So we get:
%
$$
\var{Monoidal} \simeq \var{Kleisli}
$$
%
\end{frame}
\end{document}

9
src/Cat/Category.agda
View File

@ -55,14 +55,7 @@ record RawCategory (ℓa ℓb : Level) : Set (lsuc (ℓa ⊔ ℓb)) where
-- infixl 8 _>>>_
infixl 10 _<<<_ _>>>_
-- | Operations on data
domain : {a b : Object} → Arrow a b → Object
domain {a} _ = a
codomain : {a b : Object} → Arrow a b → Object
codomain {b = b} _ = b
-- | Reverse arrow composition
_>>>_ : {A B C : Object} → (Arrow A B) → (Arrow B C) → Arrow A C
f >>> g = g <<< f

1
src/Cat/Category/Product.agda
View File

@ -1,7 +1,6 @@
{-# OPTIONS --cubical --caching #-}
module Cat.Category.Product where
open import Cat.Prelude as P hiding (_×_ ; fst ; snd)
open import Cat.Equivalence