Various changes

doc/chalmerstitle.sty

@@ -44,19 +44,16 @@
 \fontseries{sb}
 %% \fontfamily{noto}\selectfont
 {\Huge\@title}\\[.5cm]
-{\Large A formalization of category theory in Cubical Agda}\\[.5cm]
+{\Large A formalization of category theory in Cubical Agda}\\[2.5cm]
-Master's thesis in Computer Science \\[.5cm]
-{\Large\@author}\\[2cm]
 \begin{center}
 \includegraphics[width=\linewidth,keepaspectratio]{isomorphism.png}
 \end{center}
-\endgroup
 % Cover text
 \vfill
 %% \renewcommand{\familydefault}{\sfdefault} \normalfont % Set cover page font
-\textsc{Department of Computer Science and Engineering}\\
+{\Large\@author}\\[.5cm]
-\textsc{Chalmers University of Technology}\\
+Master's thesis in Computer Science
-\textsc{Gothenburg, Sweden \the\year}
+\endgroup
 %% \renewcommand{\familydefault}{\rmdefault} \normalfont % Reset standard font
 %% \end{titlepage}
 
@@ -99,7 +96,7 @@ Master's thesis in Computer Science \\[.5cm]
 \thispagestyle{plain}
 \textit{\@title}\\
 \@subtitle\\
-\copyright\ \the\year ~ \MakeUppercase{\@author}
+\copyright\ \the\year ~ \textsc{\@author}
 \vspace{4.5cm}
 
 \setlength{\parskip}{0.5cm}

doc/implementation.tex

@@ -30,60 +30,111 @@ that we can provide a proof that the laws are mere propositions. As an example a
 category is defined to have two members: `raw` which is a collection of the data
 and `isCategory` which asserts some laws about that data.
 
-This allows me to reason about things in a more mathematical way, where one can
+This allows me to reason about things in a more ``standard mathematical way'',
-reason about two categories by simply focusing on the data. This is acheived by
+where one can reason about two categories by simply focusing on the data. This
-creating a function embodying the ``equality principle'' for a given type.
+is acheived by creating a function embodying the ``equality principle'' for a
+given type.
 
 \section{Categories}
-The data for a category consist of objects, morphisms (or arrows as I will refer
+The data for a category consist of a type for the sort of objects; a type for
-to them henceforth), the identity arrow and composition of arrows.
+the sort of arrows; an identity arrow and a composition operation for arrows.
-
 Another record encapsulates some laws about this data: associativity of
 composition, identity law for the identity morphism. These are standard
-requirements for being a category as can be found in standard mathematical
+constituents of a category and can be found in typical mathematical expositions
-expositions on the topic. We, however, impose one further requirement on what it
+on the topic. We, however, impose one further requirement on what it means to be
-means to be a category, namely that the type of arrows form a set. Such
+a category, namely that the type of arrows form a set.
-categories are called \nomen{1-categories}. We could relax this requirement,
-this would give us the notion of higher categorier (\cite[p. 307]{hott-2013}).
-For the purpose of this project, however, this report will restrict itself to
-1-categories.
 
-Raw categories satisfying these properties are called a pre-categories.
+Such categories are called \nomen{1-categories}. It's possible to relax this
+requirement. This would lead to the notion of higher categorier (\cite[p.
+  307]{hott-2013}). For the purpose of this project, however, this report will
+restrict itself to 1-categories. Making based on higher categories would be a
+very natural possible extension of this work.
 
-As a further requirement to be a proper category we require it to be univalent.
+Raw categories satisfying all of the above requirements are called a
-This requirement is quite similiar to univalence for types, but we let
+\nomen{pre}-categories. As a further requirement to be a proper category we
-isomorphism on objects play the role of equivalence on types. The univalence
+require it to be univalent. Before we can define this, I must introduce two more
-criterion is:
+definitions: If we let $p$ be a witness to the identity law, which formally is:
 %
-$$
+\begin{equation}
+  \label{eq:identity}
+  \var{IsIdentity} \defeq
+  \prod_{A\ B \tp \Object} \prod_{f \tp A \to B}
+    \id \comp f \equiv f \x f \comp \id \equiv f
+\end{equation}
+%
+Then we can construct the identity isomorphism $\var{idIso} \tp \identity,
+\identity, p \tp A \approxeq A$ for any obejct $A$. Here $\approxeq$ denotes
+isomorphism on objects (whereas $\cong$ denotes isomorphism of types). This will
+be elaborated further on in sections \ref{sec:equiv} and \ref{sec:univalence}.
+Moreover, due to substitution for paths we can construct an isomorphism from
+\emph{any} path:
+%
+\begin{equation}
+\var{idToIso} : A ≡ B → A ≊ B
+\end{equation}
+%
+The univalence criterion for categories states that this map must be an
+equivalence. The requirement is similar to univalence for types, but where
+isomorphism on objects play the role of equivalence on types. Formally:
+%
+\begin{align}
+\label{eq:cat-univ}
 \isEquiv\ (A \equiv B)\ (A \approxeq B)\ \idToIso
-$$
+\end{align}
 %
-Here $\approxeq$ denotes isomorphism on objects (whereas $\cong$ denotes
+Note that \ref{eq:cat-univ} is \emph{not} the same as:
-isomorphism of types).
-
-Note that this is not the same as:
 %
-$$
+\begin{equation}
+\label{eq:cat-univalence}
+\tag{Univalence, category}
 (A \equiv B) \simeq (A \approxeq B)
-$$
+\end{equation}
 %
-The two types are logically equivalent, however. One can construct the latter
+However the two are logically equivalent: One can construct the latter from the
-from the former simply by ``forgetting'' that $\idToIso$ plays the role of the
+former simply by ``forgetting'' that $\idToIso$ plays the role of the
 equivalence. The other direction is more involved and will be discussed in
 section \ref{sec:univalence}.
 
+In summary, the definition of a category is the following collection of 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}
+%
+And laws:
+%
+\begin{align}
+\tag{associativity}
+h \lll (g \lll f) ≡ (h \lll g) \lll f \\
+\tag{identity}
+\identity \lll f ≡ f \x
+f \lll \identity ≡ f
+\\
+\label{eq:arrows-are-sets}
+\tag{arrows are sets}
+\isSet\ (\Arrow\ A\ B)\\
+\tag{\ref{eq:cat-univ}}
+\isEquiv\ (A \equiv B)\ (A \approxeq B)\ \idToIso
+\end{align}
+%
+$\lll$ denotes arrow composition (right-to-left), and reverse function
+composition (left-to-right, diagramatic order) is denoted $\rrr$. The objects
+($A$, $B$ and $C$) and arrow ($f$, $g$, $h$) are implicitly universally
+quantified.
+
 With all this in place it is now possible to prove that all the laws are indeed
 mere propositions. Most of the proofs simply use the fact that the type of
 arrows are sets. This is because most of the laws are a collection of equations
 between arrows in the category. And since such a proof does not have any content
 exactly because the type of arrows form a set, two witnesses must be the same.
-All the proofs are really quite mechanical. Lets have a look at one of them: The
+All the proofs are really quite mechanical. Lets have a look at one of them.
-identity law states that:
+Proving that \ref{eq:identity} is a mere proposition:
 %
 \begin{equation}
-  \var{IsIdentity} \defeq
+  \isProp\ \var{IsIdentity}
-  \prod_{A\ B \tp \Object} \prod_{f \tp A \to B} \id \comp f \equiv f \x f \comp \id \equiv f
 \end{equation}
 %
 There are multiple ways to prove this. Perhaps one of the more intuitive proofs
@@ -91,9 +142,9 @@ is by way of the `combinators' $\propPi$ and $\propSig$ presented in sections
 \ref{sec:propPi} and \ref{sec:propSig}:
 %
 \begin{align*}
-\mathit{propPi} & \tp \left(\prod_{a \tp A} \isProp\ (P\ a)\right) \to \isProp\ \left(\prod_{a \tp A} P\ a\right)
+\var{propPi} & \tp \left(\prod_{a \tp A} \isProp\ (P\ a)\right) \to \isProp\ \left(\prod_{a \tp A} P\ a\right)
   \\
-\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)
+\var{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{align*}
 %
 So the proof goes like this: We `eliminate' the 3 function abstractions by
@@ -103,9 +154,10 @@ $$
 \isProp \left( \id \comp f \equiv f \x f \comp \id \equiv f \right)
 $$
 %
-then we eliminate the (non-dependent) sigma-type by applying $\propSig$ giving
+Then we eliminate the (non-dependent) sigma-type by applying $\propSig$ giving
 us the two obligations: $\isProp\ (\id \comp f \equiv f)$ and $\isProp\ (f \comp
-\id \equiv f)$ which follows from the type of arrows being a set.
+\id \equiv f)$ which follows from the type of arrows being a
+set.
 
 This example illustrates nicely how we can use these combinators to reason about
 `canonical' types like $\sum$ and $\prod$. Similiar combinators can be defined
@@ -119,11 +171,14 @@ available to help us here. In stead we'll have to use the path-type directly.
 
 What we want to prove is:
 %
-$$
+\begin{equation}
+\label{eq:propIsPreCategory}
 \isProp\ \IsPreCategory
-$$
+\end{equation}
 %
-Which is judgmentally the same as
+%% \begin{proof}
+%
+This is judgmentally the same as
 %
 $$
 \prod_{a\ b \tp \IsPreCategory} a \equiv b
@@ -135,42 +190,44 @@ function $I \to \IsPreCategory$. This path must satisfy being being judgmentally
 the same as $a$ at the left endpoint and $b$ at the right endpoint. We know we
 can form a continuous path between all projections of $a$ and $b$, this follows
 from the type of all the projections being mere propositions. For instance, the
-path between $\isIdentity_a$ and $\isIdentity_b$ is simply formed by:
+path between $a.\isIdentity$ and $b.\isIdentity$ is simply formed by:
 %
 $$
-\propIsIdentity\ \isIdentity_a\ \isIdentity_b \tp \isIdentity_a \equiv \isIdentity_b
+\propIsIdentity\ a.\isIdentity\ b.\isIdentity
+\tp
+a.\isIdentity \equiv b.\isIdentity
 $$
 %
-So to give the continuous function $I \to \IsPreCategory$ that is our goal we
+So to give the continuous function $I \to \IsPreCategory$, which is our goal, we
 introduce $i \tp I$ and proceed by constructing an element of $\IsPreCategory$
-by using that all the projections are propositions to generate paths between all
+by using the fact that all the projections are propositions to generate paths
-projections. Once we have such a path e.g. $p \tp \isIdentity_a \equiv
+between all projections. Once we have such a path e.g. $p \tp a.\isIdentity
-\isIdentity_b$ we can elimiate it with $i$ and thus obtaining $p\ i \tp
+\equiv b.\isIdentity$ we can elimiate it with $i$ and thus obtain $p\ i \tp
-\isIdentity_{p\ i}$ and this element satisfies exactly that it corresponds to
+(p\ i).\isIdentity$. This element satisfies exactly that it corresponds to the
-the corresponding projections at either endpoint. Thus the element we construct
+corresponding projections at either endpoint. Thus the element we construct at
-at $i$ becomes the triple:
+$i$ becomes the triple:
 %
 \begin{equation}
 \begin{alignat}{4}
-  & \mathit{propIsAssociative} && x.\mathit{isAssociative}\
+  & \var{propIsAssociative} && a.\var{isAssociative}\
-       && y.\mathit{isAssociative} && i  \\
+       && b.\var{isAssociative} && i  \\
-  & \mathit{propIsIdentity}    && x.\mathit{isIdentity}\
+  & \var{propIsIdentity}    && a.\var{isIdentity}\
-       && y.\mathit{isIdentity}    && i  \\
+       && b.\var{isIdentity}    && i  \\
-  & \mathit{propArrowsAreSets} && x.\mathit{arrowsAreSets}\
+  & \var{propArrowsAreSets} && a.\var{arrowsAreSets}\
-       && y.\mathit{arrowsAreSets} && i
+       && b.\var{arrowsAreSets} && i
 \end{alignat}
 \end{equation}
 %
-I've found that this to be a general pattern when proving things in homotopy
+I've found this to be a general pattern when proving things in homotopy type
-type theory, namely that you have to wrap and unwrap equalities at different
+theory, namely that you have to wrap and unwrap equalities at different levels.
-levels. It is worth noting that proving this theorem with the regular inductive
+It is worth noting that proving this theorem with the regular inductive equality
-equality type would already not be possible, since we at least need
+type would already not be possible, since we at least need extensionality (the
-extensionality (the projections are all $\prod$-types). Assuming we had
+projections are all $\prod$-types). Assuming we had functional extensionality
-functional extensionality available to us as an axiom, we would use functional
+available to us as an axiom, we would use functional extensionality (in
-extensionality (in reverse?) to retreive the equalities in $a$ and $b$,
+reverse?) to retreive the equalities in $a$ and $b$, pattern-match on them to
-pattern-match on them to see that they are both $\mathit{refl}$ and then close
+see that they are both $\var{refl}$ and then close the proof with $\var{refl}$.
-the proof with $\mathit{refl}$. Of course this theorem is not so interesting in
+Of course this theorem is not so interesting in the setting of ITT since we know
-the setting of ITT since we know a priori that equality proofs are unique.
+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.
@@ -180,32 +237,35 @@ 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 homogenous:
 %
 $$
-p \tp \isPreCategory_a \equiv \isPreCategory_b
+p \tp a.\isPreCategory \equiv b.\isPreCategory
 $$
 %
 and one heterogeneous:
 %
 $$
-\Path\ (\lambda\; i \mto Univalent_{p\ i})\ \isPreCategory_a\ \isPreCategory_b
+\Path\ (\lambda\; i \to (p\ i).Univalent)\ a.\isPreCategory\ b.\isPreCategory
 $$
 %
 Which depends on the choice of $p$. The first of these we can provide since, as
-we have shown, $\IsPreCategory$ is constantly a proposition. However, even
+we have shown, $\IsPreCategory$ is a proposition. However, even though
-though $\Univalent$ is also a proposition, we cannot use this directly to show
+$\Univalent$ is also a proposition, we cannot use this directly to show the
-the latter. This is becasue $\isProp$ talks about non-dependent paths. To
+latter. This is becasue $\isProp$ talks about non-dependent paths. So we need to
-`promote' this to a dependent path we can use the combinator; $\lemPropF$
+'promote' the result that univalence is a proposition to a heterogeneous path.
-introduced in \ref{sec:lemPropF}.
+To this end we can use $\lemPropF$, which wasintroduced in \ref{sec:lemPropF}.
 
-In this case $A = \mathit{IsIdentity}\ \mathit{identity}$ and $B =
+In this case $A = \var{IsIdentity}\ \identity$ and $B = \var{Univalent}$. We've
-\mathit{Univalent}$ we've shown that being a category is a proposition, a result
+shown that being a category is a proposition, a result that holds for any choice
-that holds for any choice of identity proof. Finally we must provide a proof
+of identity proof. Finally we must provide a proof that the identity proofs at
-that the identity proofs at $a$ and $b$ are indeed the same, this we can extract
+$a$ and $b$ are indeed the same, this we can extract from $p$ by applying
-from $p$ by applying using congruence of paths:
+congruence of paths:
 %
 $$
-\congruence\ \mathit{isIdentity}\ p
+\congruence\ \var{isIdentity}\ p
 $$
 %
+And this finishes the proof that being-a-category is a mere proposition
+(\ref{eq:propIsPreCategory}).
+
 When we have a proper category we can make precise the notion of ``identifying
 isomorphic types'' \TODO{cite awodey here}. That is, we can construct the
 function:
@@ -231,25 +291,38 @@ details. \TODO{The proof is a bit fun, should I include it?}
 \label{sec:equiv}
 The usual notion of a function $f \tp A \to B$ having an inverses is:
 %
-$$
+\begin{equation}
+\label{eq:isomorphism}
 \sum_{g \tp B \to A} f \comp g \equiv \identity_{B} \x g \comp f \equiv \identity_{A}
-$$
+\end{equation}
 %
-This is defined in \cite[p. 129]{hott-2013} and is referred to as the a
+This is defined in \cite[p. 129]{hott-2013} where it is referred to as the a
-quasi-inverse. At the same place \cite{hott-2013} gives an ``interface'' for
+``quasi-inverse''. We shall refer to the type \ref{eq:isomorphism} as
-what an equivalence $\isEquiv \tp (A \to B) \to \MCU$ must supply:
+$\Isomorphism\ f$. This also gives rise to the following type:
 %
-\begin{itemize}
+\begin{equation}
-\item
+A \cong B \defeq \sum_{f \tp A \to B} \Isomorphism\ f
-  $\qinv\ f \to \isEquiv\ f$
+\end{equation}
-\item
-  $\isEquiv\ f \to \qinv\ f$
-\item
-  $\isEquiv\ f$ is a proposition
-\end{itemize}
 %
-Having such an interface gives us both 1) a way to think rather abstractly about
+At the same place \cite{hott-2013} gives an ``interface'' for what the judgment
-how to work with equivalences and 2) to use ad-hoc definitions of equivalences.
+$\isEquiv \tp (A \to B) \to \MCU$ must provide:
+%
+\begin{align}
+\var{fromIso}   & \tp \Isomorphism\ f \to \isEquiv\ f \\
+\var{toIso}     & \tp \isEquiv\ f \to \Isomorphism\ f \\
+\label{eq:propIsEquiv}
+                &\mathrel{\ } \isEquiv\ f
+\end{align}
+%
+The maps $\var{fromIso}$ and $\var{toIso}$ naturally extend to these maps:
+%
+\begin{align}
+\var{fromIsomorphism} & \tp A \cong B \to A \simeq B \\
+\var{toIsomorphism}   & \tp A \simeq B \to A \cong B
+\end{align}
+%
+Having this interface gives us both: a way to think rather abstractly about how
+to work with equivalences and a way to use ad-hoc definitions of equivalences.
 The specific instantiation of $\isEquiv$ as defined in \cite{cubical-agda} is:
 %
 $$
@@ -264,40 +337,26 @@ I give it's 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.
 
-The first function from the list of requirements we will call
+$\var{fromIso}$ can be found in \cite{cubical-agda} where it is known as
-$\mathit{fromIsomorphism}$, this is known as $\mathit{gradLemma}$ in
+$\var{gradLemma}$. The implementation of $\var{fromIso}$ as well as the proof
-\cite{cubical-agda} the second one we will refer to as $\mathit{toIsmorphism}$. It's
+that this equivalence is a proposition (\ref{eq:propIsEquiv}) can be found in my
-implementation can be found in the sources. Likewise the proof that this
+implementation.
-equivalence is propositional can be found in my implementation.
 
 We say that two types $A\;B \tp \Type$ are equivalent exactly if there exists an
 equivalence between them:
 %
-$$
+\begin{equation}
+\label{eq:equivalence}
 A \simeq B \defeq \sum_{f \tp A \to B} \isEquiv\ f
-$$
+\end{equation}
 %
 Note that the term equivalence here is overloaded referring both to the map $f
-\tp A \to B$ and the type $A \simeq B$. I will use these conflated terms when it
+\tp A \to B$ and the type $A \simeq B$. The notion of an isomorphism is
+similarly conflated as isomorphism can refer to the type $A \cong B$ as well as
+the the map $A \to B$ that witness this. I will use these conflated terms when
 it is clear from the context what is being referred to.
 
-Just like we could promote a quasi-inverse to an equivalence we can promote an
+Both $\cong$ and $\simeq$ form equivalence relations (no pun intended).
-isomorphism to an equivalence:
-%
-$$
-\mathit{fromIsomorphism} \tp A \cong B \to A \simeq B
-$$
-%
-and vice-versa:
-%
-$$
-\mathit{toIsomorphism} \tp A \simeq B \to A \cong B
-$$
-%
-The notion of an isomorphism is similarly conflated as isomorphism can refer to
-the type $A \cong B$ as well as the the map $A \to B$ that witness this.
-
-Both $\cong$ and $\simeq$ form equivalence relations.
 
 \section{Univalence}
 \label{sec:univalence}
@@ -305,7 +364,7 @@ As noted in the introduction the univalence for types $A\; B \tp \Type$ states
 that:
 %
 $$
-\mathit{Univalence} \defeq (A \equiv B) \simeq (A \simeq B)
+\var{Univalence} \defeq (A \equiv B) \simeq (A \simeq B)
 $$
 %
 As mentioned the univalence criterion for some category $\bC$ says that for all
@@ -332,9 +391,8 @@ few seconds. This type looks very similar to univalence for types and is
 therefore perhaps a bit more intuitive to grasp the implications of. Of course
 univalence for types (which is a proposition -- i.e. provable) does not imply
 univalence of all pre-category since morphisms in a category are not regular
-maps -- in stead they can be thought of as a generalization hereof; i.e. arrows.
+functions -- in stead they can be thought of as a generalization hereof. The univalence criterion therefore is simply a way of restricting arrows
-The univalence criterion therefore is simply a way of restricting arrows to
+to behave similarly to maps.
-behave similarly to maps.
 
 I will now mention a few helpful thoerems that follow from univalence that will
 become useful later.
@@ -352,66 +410,72 @@ $$
 %
 The next few theorems are variations on theorem 9.1.9 from \cite{hott-2013}. Let
 an isomorphism $A \approxeq B$ in some category $\bC$ be given. Name the
-isomorphism $\iota \tp A \to B$ and its inverse $\widetilde{\iota} \tp B \to A$.
+isomorphism $\iota \tp A \to B$ and its inverse $\inv{\iota} \tp B \to A$.
 Since $\bC$ is a category (and therefore univalent) the isomorphism induces a
 path $p \tp A \equiv B$. From this equality we can get two further paths:
-$p_{\mathit{dom}} \tp \mathit{Arrow}\ A\ X \equiv \mathit{Arrow}\ B\ X$ and
+$p_{\var{dom}} \tp \var{Arrow}\ A\ X \equiv \var{Arrow}\ B\ X$ and
-$p_{\mathit{cod}} \tp \mathit{Arrow}\ X\ A \equiv \mathit{Arrow}\ X\ B$. We
+$p_{\var{cod}} \tp \var{Arrow}\ X\ A \equiv \var{Arrow}\ X\ B$. We
 then have the following two theorems:
 %
-$$
-\mathit{coeDom} \tp \prod_{f \tp A \to X} \mathit{coe}\ p_{\mathit{dom}}\ f \equiv f \lll \widetilde{\iota}
-$$
-%
-%
 \begin{align}
+\label{eq:coeDom}
+\var{coeDom} & \tp \prod_{f \tp A \to X}
+\var{coe}\ p_{\var{dom}}\ f \equiv f \lll \inv{\iota}
+\\
 \label{eq:coeCod}
-\mathit{coeCod} \tp \prod_{f \tp A \to X} \mathit{coe}\ p_{\mathit{cod}}\ f \equiv \iota \lll f
+\var{coeCod} & \tp \prod_{f \tp A \to X}
+\var{coe}\ p_{\var{cod}}\ f \equiv \iota \lll f
 \end{align}
 %
 I will give the proof of the first theorem here, the second one is analagous.
+%
 \begin{align*}
-\mathit{coe}\ p_{\mathit{dom}}\ f
+\var{coe}\ p_{\var{dom}}\ f
-  & \equiv f \lll \mathit{obverse}_{\mathit{idToIso}\ p} && \text{lemma} \\
+  & \equiv f \lll \inv{(\var{idToIso}\ p)} && \text{lemma} \\
-  & \equiv f \lll \widetilde{\iota}
+  & \equiv f \lll \inv{\iota}
-    && \text{$\mathit{idToIso}$ and $\mathit{isoToId}$ are inverses}\\
+    && \text{$\var{idToIso}$ and $\var{isoToId}$ are inverses}\\
 \end{align*}
 %
 In the second step we use the fact that $p$ is constructed from the isomorphism
-$\iota$ -- $\mathit{obverse}$ denotes the map $B \to A$ induced by the
+$\iota$ -- $\inv{(\var{idToIso}\ p)}$ denotes the map $B \to A$ induced by the
-isomorphism $\mathit{idToIso}\ p \tp A \cong B$. The helper-lemma is similar to
+isomorphism $\var{idToIso}\ p \tp A \cong B$. The helper-lemma is similar to
 what we're trying to prove but talks about paths rather than isomorphisms:
 %
-$$
+\begin{equation}
-\prod_{f \tp \mathit{Arrow}\ A\ B} \prod_{p \tp A \equiv B} \mathit{coe}\ p_\var{dom}\ f \equiv f \lll \mathit{obverse}_{\mathit{idToIso}\ p}
+\label{eq:coeDomIso}
-$$
+\prod_{f \tp \var{Arrow}\ A\ B} \prod_{p \tp A \equiv B}
+\var{coe}\ p_\var{dom}\ f \equiv f \lll \inv{(\var{idToIso}\ p)}}
+\end{equation}
 %
-Again $p_\var{dom}$ denotes the path $\mathit{Arrow}\ A\ X \equiv
+Again $p_\var{dom}$ denotes the path $\var{Arrow}\ A\ X \equiv
-\mathit{Arrow}\ B\ X$ induced by $p$. To prove this statement I let $f$ and $p$
+\var{Arrow}\ B\ X$ induced by $p$. To prove this statement I let $f$ and $p$
 be given and then invoke based-path-induction. The induction will be based at $A
-\tp \mathit{Object}$, so let $\widetilde{A} \tp \Object$ and $\widetilde{p} \tp
+\tp \var{Object}$, so let $\widetilde{B} \tp \Object$ and $\widetilde{p} \tp
-A \equiv \widetilde{A}$ be given. The family that we perform induction over will
+A \equiv \widetilde{B}$ be given. The family that we perform induction over will
 be:
 %
 $$
-\mathit{coe}\ {\widetilde{p}}^*\ f \equiv f \lll \mathit{obverse}_{\mathit{idToIso}\ \widetilde{p}}
+\var{coe}\ {\widetilde{p}}^*\ f
+\equiv
+f \lll \inv{(\var{idToIso}\ \widetilde{p})}
 $$
 The base-case therefore becomes:
 \begin{align*}
-\mathit{coe}\ {\widetilde{\refl}}^*\ f
+\var{coe}\ {\widetilde{\refl}}^*\ f
 & \equiv f \\
-& \equiv f \lll \mathit{identity} \\
+& \equiv f \lll \var{identity} \\
-& \equiv f \lll \mathit{obverse}_{\mathit{idToIso}\ \widetilde{\refl}}
+& \equiv f \lll \inv{(\var{idToIso}\ \widetilde{\refl})}
 \end{align*}
 %
 The first step follows because reflixivity is a neutral element for coercions.
 The second step is the identity law in the category. The last step has to do
-with the fact that $\mathit{idToIso}$ is constructed by substituting according
+with the fact that $\var{idToIso}$ is constructed by substituting according to
-to the supplied path and since reflexivity is also the neutral element for
+the supplied path and since reflexivity is also the neutral element for
 substuitutions we arrive at the desired expression. To close the
-based-path-induction we must supply the value at the other end and the
+based-path-induction we must supply the value ``at the other''. In this case
-connecting path, but in this case this is simply $A' \tp \Object$ and $p \tp A
+this is simply $B \tp \Object$ and $p \tp A \equiv B$ which we have.
-\equiv A'$ which we have.
+
+And this finishes the proof of \ref{eq:coeDomIso} and thus \ref{eq:coeDom}.
 %
 \section{Categories}
 \subsection{Opposite category}
@@ -423,14 +487,18 @@ univalence in a very simple category where the structure of the category is
 rather simple.
 
 Let $\bC$ be some category, we then define the opposite category
-$\bC^{\mathit{Op}}$. It has the same objects, but the type of arrows are flipped,
+$\bC^{\var{Op}}$. It has the same objects, but the type of arrows are flipped,
 that is to say an arrow from $A$ to $B$ in the opposite category corresponds to
 an arrow from $B$ to $A$ in the underlying category. The identity arrow is the
 same as the one in the underlying category (they have the same type). Function
 composition will be reverse function composition from the underlying category.
 
-Showing that this forms a pre-category is rather straightforward. I'll state the
+I'll refer to things in terms of the underlying category, unless they have an
-laws in terms of the underlying category $\bC$:
+over-bar. So e.g. $\idToIso$ is a function in the underlying category and the
+corresponding thing is denoted $\wideoverbar{\idToIso}$ in the opposite
+category.
+
+Showing that this forms a pre-category is rather straightforward. 
 %
 $$
 h \rrr (g \rrr f) \equiv h \rrr g \rrr f
@@ -440,78 +508,77 @@ Since $\rrr$ is reverse function composition this is just the symmetric version
 of associativity.
 %
 $$
-\mathit{identity} \rrr f \equiv f \x f \rrr identity \equiv f
+\var{identity} \rrr f \equiv f \x f \rrr identity \equiv f
 $$
 %
 This is just the swapped version of identity.
 
 Finally, that the arrows form sets just follows by flipping the order of the
-arguments. Or in other words since $\Hom_{A\ B}$ is a set for all $A\ B \tp
+arguments. Or in other words; since $\Arrow\ A\ B$ is a set for all $A\;B \tp
-\Object$ then so is $\Hom_{B\ A}$.
+\Object$ then so is $Arrow\ B\ A$.
 
-Now, to show that this category is univalent is not as straight-forward. Lucliy
+Now, to show that this category is univalent is not as straight-forward. Luckily
-section \ref{sec:equiv} gave us some tools to work with equivalences.
+section \ref{sec:equiv} gave us some tools to work with equivalences. We saw
-We saw that we
+that we can prove this category univalent by giving an inverse to
-can prove this category univalent by giving an inverse to
+$\wideoverbar{\idToIso} \tp (A \equiv B) \to (A \wideoverbar{\approxeq} B)$.
-$\idToIso_{\mathit{Op}} \tp (A \equiv B) \to (A \approxeq_{\mathit{Op}} B)$.
 From the original category we have that $\idToIso \tp (A \equiv B) \to (A \cong
-B)$ is an isomorphism. Let us denote it's inverse with $\eta \tp (A \approxeq B)
+B)$ is an isomorphism. Let us denote it's inverse with $\isoToId \tp (A
-\to (A \equiv B)$. If we squint we can see what we need is a way to go between
+\approxeq B) \to (A \equiv B)$. If we squint we can see what we need is a way to
-$\approxeq_{\mathit{Op}}$ and $\approxeq$, well, an inhabitant of $A \approxeq
+go between $\wideoverbar{\approxeq}$ and $\approxeq$.
-B$ is simply an arrow $f \tp \mathit{Arrow}\ A\ B$ and it's inverse $g \tp
-\mathit{Arrow}\ B\ A$. 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 $\mathit{shuffle} \tp (A \approxeq B) \to (A
-\approxeq_{\bC} B)$ and $\mathit{shuffle}^{-1} \tp (A \approxeq_{\bC} B) \to (A
-\approxeq B)$ respectively.
 
-As the inverse of $\idToIso_{\mathit{Op}}$ I will pick $\zeta \defeq \eta \comp
+An inhabitant of $A \approxeq B$ is simply an arrow $f \tp \var{Arrow}\ A\ B$
-\var{shuffle}$. The proof that they are inverses go as follows:
+and it's inverse $g \tp \var{Arrow}\ B\ A$. 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 $\var{shuffle} \tp (A \approxeq
+B) \to (A \wideoverbar{\approxeq} B)$ and $\var{shuffle}^{-1} \tp (A
+\wideoverbar{\approxeq} B) \to (A \approxeq B)$ respectively.
+
+As the inverse of $\wideoverbar{\idToIso}$ I will pick $\wideoverbar{\isoToId}
+\defeq \isoToId \comp \var{shuffle}$. The proof that they are inverses go as
+follows:
 %
 \begin{align*}
-\zeta \comp \idToIso & \equiv
+\wideoverbar{\isoToId} \comp \wideoverbar{\idToIso} & =
-\eta \comp \var{shuffle} \comp \idToIso
+\isoToId \comp \var{shuffle} \comp \wideoverbar{\idToIso}
-&& \text{by definition} \\
+\\
 %% ≡⟨ cong (λ φ → φ x) (cong (λ φ → η ⊙ shuffle ⊙ φ) (funExt lem)) ⟩ \\
 %
 & \equiv
-\eta \comp \var{shuffle} \comp \inv{\var{shuffle}} \comp \idToIso
+\isoToId \comp \var{shuffle} \comp \inv{\var{shuffle}} \comp \idToIso
 && \text{lemma} \\
 %%   ≡⟨⟩ \\
 & \equiv
-\eta \comp \idToIso_{\bC}
+\isoToId \comp \idToIso
 && \text{$\var{shuffle}$ is an isomorphism} \\
-%%   ≡⟨ (λ i → verso-recto i x) ⟩ \\
 & \equiv
 \identity
-&& \text{$\eta$ is an ismorphism}
+&& \text{$\isoToId$ is an ismorphism}
 \end{align*}
 %
 The other direction is analogous.
 
-The lemma used in this proof states that $\idToIso \equiv \inv{\var{shuffle}} \comp
+The lemma used in step 2 of this proof states that $\wideoverbar{idToIso} \equiv
-\idToIso_{\bC}$ it's a rather straight-forward proof since being-an-inverse-of
+\inv{\var{shuffle}} \comp \idToIso$. This is a rather straight-forward proof
-is a proposition.
+since being-an-inverse-of is a proposition, so it suffices to show that their
-
+first components are equal, but this holds judgmentally.
-So, in conclusion, we've shown that the opposite category is indeed a category.
 
 This finished the proof that the opposite category is in fact a category. Now,
 to prove that that opposite-of is an involution we must show:
 %
 $$
-\prod_{\bC \tp \mathit{Category}} \left(\bC^{\mathit{Op}}\right)^{\mathit{Op}} \equiv \bC
+\prod_{\bC \tp \var{Category}} \left(\bC^{\var{Op}}\right)^{\var{Op}} \equiv \bC
 $$
 %
-As we've seen the laws in $\left(\bC^{\mathit{Op}}\right)^{\mathit{Op}}$ get
+As we've seen the laws in $\left(\bC^{\var{Op}}\right)^{\var{Op}}$ get quite
-quite involved.\footnote{We haven't even seen the full story because we've used
+involved.\footnote{We haven't even seen the full story because we've used this
-  this `interface' for equivalences.} Luckily since being-a-category is a mere
+  `interface' for equivalences.} Luckily since being-a-category is a mere
 proposition, we need not concern ourselves with this bit when proving the above.
-We can use the equality principle for categories that lets us prove an equality
+We can use the equality principle for categories that let us prove an equality
 just by giving an equality on the data-part. So, given a category $\bC$ all we
 must provide is the following proof:
 %
 $$
-\mathit{raw}\ \left(\bC^{\mathit{Op}}\right)^{\mathit{Op}} \equiv \mathit{raw}\ \bC
+\var{raw}\ \left(\bC^{\var{Op}}\right)^{\var{Op}} \equiv \var{raw}\ \bC
 $$
 %
 And these are judgmentally the same. I remind the reader that the left-hand side
@@ -521,23 +588,23 @@ is constructed by flipping the arrows, which judgmentally is an involution.
 The category of sets has as objects, not types, but only those types that are
 homotopic sets. This is encapsulated in Agda with the following type:
 %
-$$\Set_\ell \defeq \sum_{A \tp \MCU_\ell} \isSet\ A$$
+$$\Set \defeq \sum_{A \tp \MCU} \isSet\ A$$
 %
 The more straight-forward notion of a category where the objects are types is
-not a valid (1-)category. Since in cubical Agda types can have higher homotopic
+not a valid \mbox{(1-)category}. This stems from the fact that types in cubical
-structure.
+Agda types can have higher homotopic structure.
 
 Univalence does not follow immediately from univalence for types:
 %
 $$(A \equiv B) \simeq (A \simeq B)$$
 %
 Because here $A\ B \tp \Type$ whereas the objects in this category have the type
-$\Set$ so we cannot form the type $\mathit{hA} \simeq \mathit{hB}$ for objects
+$\Set$ so we cannot form the type $\var{hA} \simeq \var{hB}$ for objects
-$\mathit{hA}\;\mathit{hB} \tp \Set$. In stead I show that this category
+$\var{hA}\;\var{hB} \tp \Set$. In stead I show that this category
 satisfies:
 %
 $$
-(\mathit{hA} \equiv \mathit{hB}) \simeq (\mathit{hA} \approxeq \mathit{hB})
+(\var{hA} \equiv \var{hB}) \simeq (\var{hA} \approxeq \var{hB})
 $$
 %
 Which, as we saw in section \ref{sec:univalence}, is sufficient to show that the
@@ -561,7 +628,7 @@ together. Step one will be proven with the following lemma:
 \end{align}
 %
 The lemma states that for pairs whose second component are mere propositions
-equiality is equivalent to equality of the first components. In this case the
+equality is equivalent to equality of the first components. In this case the
 type-family $P$ is $\isSet$ which itself is a proposition for any type $A \tp
 \Type$. Step two is univalence. Step three will be proven with the following
 lemma:
@@ -574,29 +641,31 @@ lemma:
 Which says that if two type-families are equivalent at all points, then pairs
 with identitical first components and these families as second components will
 also be equivalent. For our purposes $P \defeq \isEquiv\ A\ B$ and $Q \defeq
-\mathit{Isomorphism}$. So we must finally prove:
+\var{Isomorphism}$. So we must finally prove:
 %
 \begin{align}
   \label{eq:equivIso}
-\prod_{f \tp A \to B} \left( \isEquiv\ A\ B\ f \simeq \mathit{Isomorphism}\ f \right)
+\prod_{f \tp A \to B} \left( \isEquiv\ A\ B\ f \simeq \var{Isomorphism}\ f \right)
 \end{align}
 
 First, lets proove \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 $\mathit{fromIsomorphism}$ it suffices to give an isomorphism between
+of $\var{fromIsomorphism}$ it suffices to give an isomorphism between
 $x \equiv y$ and $\fst\ x \equiv \fst\ y$:
 %
-\begin{align*}
+\begin{alignat*}{5}
-  f & \defeq \congruence\ \fst \tp x \equiv y \to \fst\ x \equiv \fst\ y \\
+  f & \defeq \congruence\ \fst
-  g & \defeq \mathit{lemSig}\ \mathit{propP}\ x\ y \tp \fst\ x \equiv \fst\ y \to x \equiv y
+    && \tp x       && \equiv y       && \to \fst\ x && \equiv \fst\ y \\
-\end{align*}
+  g & \defeq \var{lemSig}\ \var{propP}\ x\ y
+    && \tp \fst\ x && \equiv \fst\ y && \to x       && \equiv y
+\end{alignat*}
 %
-\TODO{Is it confusing that I use point-free style here?}
+\TODO{Is it confusing that I use point-free style here?} Here $\var{lemSig}$ is
-Here $\mathit{lemSig}$ is a lemma that says that if the second component of a
+a lemma that says that if the second component of a pair is a proposition, it
-pair is a proposition, it suffices to give a path between it's first components
+suffices to give a path between its first components to construct an equality of
-to construct an equality of the two pairs:
+the two pairs:
 %
 \begin{align*}
-\mathit{lemSig} \tp \left( \prod_{x \tp A} \isProp\ (B\ x) \right) \to
+\var{lemSig} \tp \left( \prod_{x \tp A} \isProp\ (B\ x) \right) \to
 \prod_{u\; v \tp \sum_{a \tp A} B\ a}
   \left( \fst\ u \equiv \fst\ v \right) \to u \equiv v
 \end{align*}
@@ -615,31 +684,31 @@ Lastly we prove \ref{eq:equivIso}. Let $f \tp A \to B$ be given. For the maps we
 choose:
 %
 \begin{align*}
-\mathit{toIso}
+\var{toIso}
-  & \tp \isEquiv\ f             \to \mathit{Isomorphism}\ f \\
+  & \tp \isEquiv\ f             \to \var{Isomorphism}\ f \\
-\mathit{fromIso}
+\var{fromIso}
-  & \tp \mathit{Isomorphism}\ f \to \isEquiv\ f
+  & \tp \var{Isomorphism}\ f \to \isEquiv\ f
 \end{align*}
 %
 As mentioned in section \ref{sec:equiv}. These maps are not in general inverses
 of each other. In stead, we will use the fact that $A$ and $B$ are sets. The first thing we must prove is:
 %
 \begin{align*}
-  \mathit{fromIso} \comp \mathit{toIso} \equiv \identity_{\isEquiv\ f}
+  \var{fromIso} \comp \var{toIso} \equiv \identity_{\isEquiv\ f}
 \end{align*}
 %
 For this we can use the fact that being-an-equivalence is a mere proposition.
 For the other direction:
 %
 \begin{align*}
-  \mathit{toIso} \comp \mathit{fromIso} \equiv \identity_{\mathit{Isomorphism}\ f}
+  \var{toIso} \comp \var{fromIso} \equiv \identity_{\var{Isomorphism}\ f}
 \end{align*}
 %
-We will show that $\mathit{Isomorphism}\ f$ is also a mere proposition. To this
+We will show that $\var{Isomorphism}\ f$ is also a mere proposition. To this
-end, let $X\;Y \tp \mathit{Isomorphism}\ f$ be given. Name the maps $x\;y \tp B
+end, let $X\;Y \tp \var{Isomorphism}\ f$ be given. Name the maps $x\;y \tp B
 \to A$ respectively. Now, the proof that $X$ and $Y$ are the same is a pair of
-paths: $p \tp x \equiv y$ and $\Path\ (\lambda\; i \mto
+paths: $p \tp x \equiv y$ and $\Path\ (\lambda\; i \to
-\mathit{AreInverses}\ f\ (p\ i))\ \mathcal{X}\ \mathcal{Y}$ where $\mathcal{X}$
+\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:
 %
@@ -659,10 +728,10 @@ proposition and then use $\lemPropF$. So we prove the generalization:
 %
 \begin{align}
 \label{eq:propAreInversesGen}
-\prod_{g \tp B \to A} \isProp\ (\mathit{AreInverses}\ f\ g)
+\prod_{g \tp B \to A} \isProp\ (\var{AreInverses}\ f\ g)
 \end{align}
 %
-But $\mathit{AreInverses}\ f\ g$ is a pair of equations on arrows, so we use
+But $\var{AreInverses}\ f\ g$ is a pair of equations on arrows, so we use
 $\propSig$ and the fact that both $A$ and $B$ are sets to close this proof.
 
 \subsection{Category of categories}
@@ -676,26 +745,26 @@ These lemmas can be used to provide the actual exponential object in a context
 where we have a witness to this being a category. This is useful if this library
 is later extended to talk about higher categories.
 
-\section{Product}
+\section{Products}
 In the following I'll demonstrate a technique for using categories to prove
 properties. The goal in this section is to show that products are propositions:
 %
 $$
-\prod_{\bC \tp \Category} \prod_{A\;B \tp \Object} \isProp\ (\mathit{Product}\ \bC\ A\ B)
+\prod_{\bC \tp \Category} \prod_{A\;B \tp \Object} \isProp\ (\var{Product}\ \bC\ A\ B)
 $$
 %
-Where $\mathit{Product}\ \bC\ A\ B$ denotes the type of products of objects $A$
+Where $\var{Product}\ \bC\ A\ B$ denotes the type of products of objects $A$
 and $B$ in the category $\bC$. I do this by constructing a category whose
 terminal objects are equivalent to products in $\bC$, and since terminal objects
 are propositional in a proper category and equivalences preserve homotopy level,
 then we know that products also are propositions. But before we get to that,
 let's recall the definition of products.
 
-\subsection{Products}
+\subsection{Definition of products}
 Given a category $\bC$ and two objects $A$ and $B$ in $\bC$ we define the
-product of $A$ and $B$ to be an object $A \x B$ in $\bC$ and two arrows $\pi_1
+product (object) of $A$ and $B$ to be an object $A \x B$ in $\bC$ and two arrows
-\tp A \x B \to A$ and $\pi_2 \tp A \x B \to B$ called the projections of the
+$\pi_1 \tp A \x B \to A$ and $\pi_2 \tp A \x B \to B$ called the projections of
-product. The projections must satisfy the following property:
+the product. The projections must satisfy the following property:
 
 For all $X \tp Object$, $f \tp \Arrow\ X\ A$ and $g \tp \Arrow\ X\ B$ we have
 that there exists a unique arrow $\pi \tp \Arrow\ X\ (A \x B)$ satisfying
@@ -1017,7 +1086,7 @@ 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{$\mathit{coeDom}$ and the isomorphism $f, \inv{f}$} \\
+  & ≡ x_\mathcal{A} \lll \fst\ \inv{f} && \text{$\var{coeDom}$ and the isomorphism $f, \inv{f}$} \\
   & ≡ y_\mathcal{A} && \text{\ref{eq:pairArrowLaw} for $\inv{f}$}
 \end{align*}
 %
@@ -1214,7 +1283,5 @@ And likewise in the kleisli formulation we can define $\join$:
 \join \defeq \bind\ \identity
 \end{align}
 %
-Which shall play the role of $\join$.
-
 It now remains to show that we can prove the various laws given this choice. I
 refer the reader to my implementation for the details.

doc/introduction.tex

@@ -24,7 +24,7 @@ $f \defeq (n : \bN) \mapsto (0 + n : \bN)$
 $g \defeq (n : \bN) \mapsto (n + 0 : \bN)$
 \end{multicols}
 %
-$n + 0$ is \nomen{definitionally} equal to $n$ which we write as $n + 0 = n$.
+$n + 0$ is \nomen{definitionally} equal to $n$, which we write as $n + 0 = n$.
 This is also called \nomen{judgmental} equality. We call it definitional
 equality because the \emph{equality} arises from the \emph{definition} of $+$
 which is:
@@ -39,7 +39,7 @@ which is:
 Note that $0 + n$ is \emph{not} definitionally equal to $n$. $0 + n$ is in
 normal form. I.e.; there is no rule for $+$ whose left-hand-side matches this
 expression. We \emph{do}, however, have that they are \nomen{propositionally}
-equal which we write as $n + 0 \equiv n$. Propositional equality means that
+equal, which we write as $n + 0 \equiv n$. Propositional equality means that
 there is a proof that exhibits this relation. Since equality is a transitive
 relation we have that $n + 0 \equiv 0 + n$.
 
@@ -62,7 +62,7 @@ for a category $\bC$ and a fixed object in $A \in \bC$ is defined to be:
 \end{align*}
 %
 The proof obligation that this satisfies the identity law of functors
-($\fmap\ \idFun \equiv \idFun$) becomes:
+($\fmap\ \idFun \equiv \idFun$) thus becomes:
 %
 \begin{align*}
 \Hom(A, \idFun_{\bX}) = (g \mapsto \idFun \comp g) \equiv \idFun_{\Sets}
@@ -86,7 +86,7 @@ immediately conclude $\{x \mid \phi\ x \land \psi\ x\} \equiv \{x \mid
 \phi\ x\}$ without thinking twice. Unfortunately such an identification can not
 be performed in ITT.
 
-More specifically; what we are interested in is a way of identifying
+More specifically what we are interested in is a way of identifying
 \nomen{equivalent} types. I will return to the definition of equivalence later
 in section \ref{sec:equiv}, but for now it is sufficient to think of an
 equivalence as a one-to-one correspondence. We write $A \simeq B$ to assert that
@@ -102,7 +102,7 @@ In particular this allows us to construct an equality from an equivalence
 The above examples serve to illustrate a limitation of ITT. One case where these
 limitations are particularly prohibitive is in the study of Category Theory. At
 a glance category theory can be described as ``the mathematical study of
-(abstract) algebras of functions'' (\cite{awodey-2006}). So by that token
+(abstract) algebras of functions'' (\cite{awodey-2006}). By that token
 functional extensionality is particularly useful for formulating Category
 Theory. In Category theory it is also common to identify isomorphic structures
 and univalence gives us a way to make this notion precise. In fact we can
@@ -111,28 +111,38 @@ formulate this requirement within our formulation of categories by requiring the
 
 \section{Context}
 %
-\begin{verbatim}
-Inspiration:
-* Awodey - formulation of categories
-* HoTT   - sketch of homotopy proofs
-\end{verbatim}
 The idea of formalizing Category Theory in proof assistants is not new. There
 are a multitude of these available online. Just as a first reference see this
 question on Math Overflow: \cite{mo-formalizations}. Notably these
 implementations of category theory in Agda:
+%
 \begin{itemize}
 \item
-\url{https://github.com/copumpkin/categories} -- setoid interpretation
+  \url{https://github.com/copumpkin/categories}
-\item
-\url{https://github.com/pcapriotti/agda-categories} -- homotopic setting with postulates
-\item
-\url{https://github.com/pcapriotti/agda-categories} -- homotopic setting in coq
-\item
-\url{https://github.com/mortberg/cubicaltt} -- homotopic setting in \texttt{cubicaltt}
-\end{itemize}
 
+  A formalization in Agda using the setoid approach
+\item
+  \url{https://github.com/pcapriotti/agda-categories}
+
+  A formalization in Agda with univalence and functional extensionality as
+  postulates.
+\item
+  \url{https://github.com/pcapriotti/agda-categories}
+
+  A formalization in Coq in the homotopic setting
+\item
+  \url{https://github.com/mortberg/cubicaltt}
+
+  A formalization in CubicalTT - a language designed for cubical-type-theory.
+  Formalizes many different things, but only a few concepts from category
+  theory.
+
+\end{itemize}
+%
 The contribution of this thesis is to explore how working in a cubical setting
-will make it possible to prove more things and to reuse proofs.
+will make it possible to prove more things and to reuse proofs and to try and
+compare some aspects of this formalization with the existing ones.\TODO{How can
+  I live up to this?}
 
 There are alternative approaches to working in a cubical setting where one can
 still have univalence and functional extensionality. One option is to postulate
@@ -147,11 +157,44 @@ canonical form.
 Another approach is to use the \emph{setoid interpretation} of type theory
 (\cite{hofmann-1995,huber-2016}). With this approach one works with
 \nomen{extensionals sets} $(X, \sim)$, that is a type $X \tp \MCU$ and an
-equivalence relation $\sim \tp X \to X \to \MCU$.
+equivalence relation $\sim \tp X \to X \to \MCU$ on that type. Under the setoid
-
+interpretation the equivalence relation serve as a sort of ``local''
-Types should additionally `carry around' an equivalence relation that serve as
 propositional equality. This approach has other drawbacks; it does not satisfy
-all judgemental equalites of type theory, is cumbersome to work with in practice
+all propositional equalites of type theory (\TODO{Citation needed}, is
-(\cite[p. 4]{huber-2016}) and makes equational proofs less reusable since
+cumbersome to work with in practice (\cite[p. 4]{huber-2016}) and makes
-equational proofs $a \sim_{X} b$ are inherently `local' to the extensional set
+equational proofs less reusable since equational proofs $a \sim_{X} b$ are
-$(X , \sim)$.
+inherently `local' to the extensional set $(X , \sim)$.
+
+\section{Conventions}
+\TODO{Talk a bit about terminology. Find a good place to stuff this little
+  section.}
+
+In the remainder of this paper I will use the term \nomen{Type} to describe --
+well, types. Thereby diverging from the notation in Agda where the keyword
+\texttt{Set} refers to types. \nomen{Set} on the other hand shall refer to the
+homotopical notion of a set. I will also leave all universe levels implicit.
+
+And I use the term \nomen{arrow} to refer to morphisms in a category, whereas
+the terms morphism, map or function shall be reserved for talking about
+type-theoretic functions; i.e. functions in Agda.
+
+$\defeq$ will be used for introducing definitions. $=$ will be used to for
+judgmental equality and $\equiv$ will be used for propositional equality.
+
+All this is summarized in the following table:
+
+\begin{center}
+\begin{tabular}{ c c c }
+Name & Agda & Notation \\
+\hline
+\nomen{Type}            & \texttt{Set}         & $\Type$            \\
+\nomen{Set}             & \texttt{Σ Set IsSet} & $\Set$             \\
+Function, morphism, map & \texttt{A → B}       & $A → B$            \\
+Dependent- ditto        & \texttt{(a : A) → B} & $∏_{a \tp A} → B$  \\
+\nomen{Arrow}           & \texttt{Arrow A B}   & $\Arrow\ A\ B$     \\
+\nomen{Object}          & \texttt{C.Object}    & $̱ℂ.Object$
+Definition              & \texttt{=}           & $̱\defeq$
+Judgmental equality     & \null                & $̱=$
+Propositional equality  & \null                & $̱\equiv$
+\end{tabular}
+\end{center}

doc/macros.tex

@@ -71,6 +71,7 @@
 \newcommand\Univalent{\var{Univalent}}
 \newcommand\refl{\var{refl}}
 \newcommand\isoToId{\var{isoToId}}
+\newcommand\Isomorphism{\var{Isomorphism}}
 \newcommand\rrr{\ggg}
 \newcommand\fish{\mathrel{\wideoverbar{\rrr}}}
 \newcommand\fst{\var{fst}}

doc/main.tex

@@ -44,6 +44,10 @@
 \bibliographystyle{plain}
 
 \addtocontents{toc}{\protect\thispagestyle{empty}}
+%% \newtheorem{prop}{Proposition}
+\makeatletter
+\newcommand*{\rom}[1]{\expandafter\@slowroman\romannumeral #1@}
+\makeatother
 \begin{document}
 
 \pagenumbering{roman}

doc/refs.bib

@@ -106,11 +106,11 @@
 @MISC{mo-formalizations,
     TITLE = {Formalizations of category theory in proof assistants},
     AUTHOR = {Jason Gross},
-    HOWPUBLISHED = {MathOverflow},
     NOTE = {Version: 2014-01-19},
     year={2014},
     EPRINT = {\url{https://mathoverflow.net/q/152497}},
-    URL = {https://mathoverflow.net/q/152497}
+    url = {https://mathoverflow.net/q/152497},
+    howpublished = {MathOverflow: \url{https://mathoverflow.net/q/152497}}
 }
 @Misc{UniMath,
     author = {Voevodsky, Vladimir and Ahrens, Benedikt and Grayson, Daniel and others},