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Finish section on category of sets

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Frederik Hanghøj Iversen 2018-04-24 14:11:22 +02:00
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@ -0,0 +1,72 @@
Andrea Vezzosi <vezzosi@chalmers.se> Tue, Apr 24, 2018 at 2:02 PM
To: Frederik Hanghøj Iversen <fhi.1990@gmail.com>
Cc: Thierry Coquand <coquand@chalmers.se>
On Tue, Apr 24, 2018 at 12:57 PM, Frederik Hanghøj Iversen
<fhi.1990@gmail.com> wrote:
> I've written the first few sections about my implementation. I was wondering
> if you could have a quick look at it. You don't need to read it
> word-for-word but I would like some indication from you if this is the sort
> of thing you would like to see in the final report.
Yes! I would say this very much fits the bill of what the main part of
the report should be, then you could have a discussion section where
you might put some analysis of the pros and cons of cubical, design
choices you made, and your experience overall.
I wonder if there should be some short introduction to Cubical Type
Theory before this chapter, so you can introduce the Path type by
itself and show some simple proof with it. e.g. how to get function
extensionality.
You mention a few "combinators" like propPi and lemPropF, you might
want to call them just lemmas, so it's clearer that these can be
proven in --cubical.
>
> I refer you specifically to "Chapter 2 - Implementation" on p. 6.
>
> In this chapter I plan to additionally include some text about the proof we
> did that products are mere propositions and the proof about the two
> equivalent notions of a monad.
I've read the chapter up until 2.3 and skimmed the rest for now, but I
accumulated some editing suggestions I copy here.
Remember to look for things like these when you proof-read the rest :)
You should be careful to properly introduce things before you use
them, like IsPreCategory (I'd prefer if it took the raw category as
argument btw) and its fields isIdentity, isAssociative, .. come up a
bit out of the blue from the end of page 8.
Maybe the easiest is to show the definition of IsPreCategory.
Maybe give a type for propIsIdentity and mention the other prop* are similar.
Also the notation "isIdentity_a" to apply projections is a bit unusual
so it needs to be introduced as well.
To be fair it would be simpler to stick to function application
(though I see that it would introduce more parentheses),
"The situation is a bit more complicated when we have a dependent
type" could be more clear by being more specific:
"The situation is a bit more complicated when the type of a field
depends on a previous field"
Here too it might be more concrete if you also give the code for IsCategory.
In Path ( λ i → Univalent_{p i} ) isPreCategory_a isPreCategory_b
I suggest parentheses around (p i), but also you should be consistent
on whether you want to call the proof "p" or "p_{isPreCategory}",
finally i'm guessing the two fields should be "isUnivalent" rather
than "isPreCategory".
You can cite the book on the specific definition of isEquiv,
"contractible fibers" in section 4.4, the grad lemma is also from
somewhere but I don't remember off-hand.
You have not defined what you mean by _\~=_ and isomorphism.
Cheers,
Andrea
[Quoted text hidden]

169
doc/implementation.tex
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@ -189,7 +189,7 @@ $$
and one heterogeneous:
%
$$
Path\ (\lambda i \to Univalent_{p\ i})\ \isPreCategory_a\ \isPreCategory_b
\Path\ (\lambda\; i \mto Univalent_{p\ i})\ \isPreCategory_a\ \isPreCategory_b
$$
%
Which depends on the choice of $p_{\isPreCategory}$. The first of these we can
@ -203,7 +203,7 @@ path between some two elements in $A$; $p : a_0 \equiv a_1$ then we can built a
heterogeneous path between any two $b$'s at the endpoints:
%
$$
Path\ (\lambda i \to B\ (p\ i))\ b0\ b1
\Path\ (\lambda\; i \mto B\ (p\ i))\ b0\ b1
$$
%
where $b_0 \tp B a_0$ and $b_1 \tp B\ a_1$. This is quite a mouthful, but the
@ -216,7 +216,7 @@ applying using congruence of paths: $\congruence\ \mathit{isIdentity}\
p_{\isPreCategory}$
When we have a proper category we can make precise the notion of ``identifying
isomorphic types'' (TODO cite awodey here). That is, we can construct the
isomorphic types'' \TODO{cite awodey here}. That is, we can construct the
function:
%
$$
@ -227,11 +227,11 @@ One application of this, and a perhaps somewhat surprising result, is that
terminal objects are propositional. Intuitively; they do not have any
interesting structure. The proof of this follows from the usual observation that
any two terminal objects are isomorphic. The proof is omitted here, but the
curious reader can check the implementation for the details. (TODO: The proof is
a bit fun, should I include it?)
curious reader can check the implementation for the details. \TODO{The proof is
a bit fun, should I include it?}
\section{Equivalences}
\label{equiv}
\label{sec:equiv}
The usual notion of a function $f : A \to B$ having an inverses is:
%
$$
@ -253,7 +253,7 @@ what an equivalence $\isEquiv : (A \to B) \to \MCU$ must supply:
%
Having such an interface gives us both 1) a way to think rather abstractly about
how to work with equivalences and 2) to use ad-hoc definitions of equivalences.
The specific instantiation of $\isEquiv$ as defined in \cite{cubical} is:
The specific instantiation of $\isEquiv$ as defined in \cite{cubical-agda} is:
%
$$
isEquiv\ f \defeq \prod_{b : B} \isContr\ (\fiber\ f\ b)
@ -269,7 +269,7 @@ once we have shown that this definition actually works as an equivalence.
The first function from the list of requirements we will call
$\mathit{fromIsomorphism}$, this is known as $\mathit{gradLemma}$ in
\cite{cubical} the second one we will refer to as $\mathit{toIsmorphism}$. It's
\cite{cubical-agda} the second one we will refer to as $\mathit{toIsmorphism}$. It's
implementation can be found in the sources. Likewise the proof that this
equivalence is propositional can be found in my implementation.
@ -353,7 +353,7 @@ $$
\isoToId \tp A \approxeq B \to A \equiv B
$$
%
The next few theorems are variations on theorem 9.1.9 from \cite{HoTT-book}. Let
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 $\widetilde{\iota} \tp B \to A$.
Since $\bC$ is a category (and therefore univalent) the isomorphism induces a
@ -425,7 +425,7 @@ 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
$\bC^{\matit{Op}}$. It has the same objects, but the type of arrows are flipped,
$\bC^{\mathit{Op}}$. 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
@ -442,7 +442,7 @@ Since $\rrr$ is reverse function composition this is just the symmetric version
of associativity.
%
$$
\matit{identity} \rrr f \equiv f \x f \rrr identity \equiv f
\mathit{identity} \rrr f \equiv f \x f \rrr identity \equiv f
$$
%
This is just the swapped version of identity.
@ -452,7 +452,7 @@ arguments. Or in other words since $\Hom_{A\ B}$ is a set for all $A\ B \tp
\Object$ then so is $\Hom_{B\ A}$.
Now, to show that this category is univalent is not as straight-forward. Lucliy
section \ref{equiv} gave us some tools to work with equivalences. We saw that we
section \ref{sec:equiv} gave us some tools to work with equivalences. We saw that we
can prove this category univalent by giving an inverse to
$\idToIso_{\mathit{Op}} \tp (A \equiv B) \to (A \approxeq_{\mathit{Op}} B)$.
From the original category we have that $\idToIso \tp (A \equiv B) \to (A \cong
@ -500,7 +500,7 @@ 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:
%
$$
\prod_{\bC \tp \mathit{Category}} \left(\bC^{\matit{Op}}\right)^{\matit{Op}} \equiv \bC
\prod_{\bC \tp \mathit{Category}} \left(\bC^{\mathit{Op}}\right)^{\mathit{Op}} \equiv \bC
$$
%
As we've seen the laws in $\left(\bC^{\mathit{Op}}\right)^{\mathit{Op}}$ get
@ -543,20 +543,23 @@ $$
%
Which, as we saw in section \ref{univalence}, is sufficient to show that the
category is univalent. The way that I have shown this is with a three-step
process. For objects $(A, s_A)\; (B, s_B) \tp \Set$ I show that.
process. For objects $(A, s_A)\; (B, s_B) \tp \Set$ I show the following chain
of equivalences:
%
\begin{align*}
((A, s_A) \equiv (B, s_B)) & \simeq (A \equiv B) \\
(A \equiv B) & \simeq (\fst A \simeq \fst B) \\
(A \simeq B) & \simeq ((A, s_A) \approxeq (B, s_B))
((A, s_A) \equiv (B, s_B))
& \simeq (A \equiv B) && \ref{eq:equivPropSig} \\
& \simeq (A \simeq B) && \text{Univalence} \\
& \simeq ((A, s_A) \approxeq (B, s_B)) && \text{\ref{eq:equivSig} and \ref{eq:equivIso}}
\end{align*}
And since $\simeq$ is an equivalence relation we can chain these equivalences
together. Step one will be proven with the following lemma:
%
$$
\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)
$$
\end{align}
%
The lemma states that for pairs whose second component are mere propositions
equiality is equivalent to equality of the first components. In this case the
@ -564,21 +567,105 @@ type-family $P$ is $\isSet$ which itself is a proposition for any type $A \tp
\Type$. Step two is univalence. Step three will be proven with the following
lemma:
%
$$
\begin{align}
\label{eq:equivSig}
\prod_{a \tp A} \left( P\ a \simeq Q\ a \right) \to \sum_{a \tp A} P\ a \simeq \sum_{a \tp A} Q\ a
$$
\end{align}
%
Which says that if two type-families are equivalent at all points, then pairs
with identitical first components and these families as second components will
also be equivalent. For our purposes $P \defeq \isEquiv\ A\ B$ and $Q \defeq
\mathit{Isomorphism}$. So we must finally prove:
%
$$
\begin{align}
\label{eq:equivIso}
\prod_{f \tp A \to B} \left( \isEquiv\ A\ B\ f \simeq \mathit{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
$x \equiv y$ and $\fst\ x \equiv \fst\ y$:
%
\begin{align*}
f & \defeq \congruence\ \fst \tp x \equiv y \to \fst\ x \equiv \fst\ y \\
g & \defeq \mathit{lemSig}\ \mathit{propP}\ x\ y \tp \fst\ x \equiv \fst\ y \to x \equiv y
\end{align*}
%
\TODO{Is it confusing that I use point-free style here?}
Here $\mathit{lemSig}$ is a lemma that says that if the second component of a
pair is a proposition, it suffices to give a path between it's first components
to construct an equality of the two pairs:
%
\begin{align*}
\mathit{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*}
%
The proof that these are indeed inverses has been omitted. \TODO{Do I really
want to ommit it?}\QED
\subsection{Categories}
Now to prove \ref{eq:equivSig}: Let $e \tp \prod_{a \tp A} \left( P\ a \simeq
Q\ a \right)$ be given. To prove the equivalence, it suffices to give an
isomorphism between $\sum_{a \tp A} P\ a$ and $\sum_{a \tp A} Q\ a$, but since
they have identical first components it suffices to give an isomorphism between
$P\ a$ and $Q\ a$ for all $a \tp A$. This is exactly what we can get from
the equivalence $e$.\QED
Lastly we prove \ref{eq:equivIso}. Let $f \tp A \to B$ be given. For the maps we
choose:
%
\begin{align*}
\mathit{toIso}
& \tp \isEquiv\ f \to \mathit{Isomorphism}\ f \\
\mathit{fromIso}
& \tp \mathit{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}
\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}
\end{align*}
%
We will show that $\mathit{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
\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
\mathit{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:
%
\begin{align*}
x
& \equiv x \comp \identity \\
& \equiv x \comp (f \comp y)
&& \text{$y$ is an inverse to $f$} \\
& \equiv (x \comp f) \comp y \\
& \equiv \identity \comp y
&& \text{$x$ is an inverse to $f$} \\
& \equiv y
\end{align*}
%
For the other (dependent) path we can prove that being-an-inverse-of is a
proposition and then use $\lemPropF$. So we prove the generalization:
%
\begin{align*}
\prod_{g : B \to A} \isProp\ (\mathit{AreInverses}\ f\ g)
\end{align*}
%
But $\mathit{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}
Note that this category does in fact not exist. In stead I provide the
definition of the ``raw'' category as well as some of the laws.
@ -589,8 +676,40 @@ 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}
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)
$$
%
Where $\mathit{Product}\ \bC\ A\ B$ 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 preservehomotopy level,
then we know that products also are propositions. But before we get to that,
let's recall the 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 $\pi_1 \tp A \x
B \to A$ and $\pi_2 \tp A \x B \to B$ called the projections of 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
%
\begin{align}
\label{eq:umpProduct}
%% \prod_{X \tp Object} \prod_{f \tp \Arrow\ X\ A} \prod_{g \tp \Arrow\ X\ B}\\
%% \uexists_{f \x g \tp \Arrow\ X\ (A \x B)}
\pi_1 \lll \pi \equiv f \x \pi_2 \lll \pi \equiv g
%% ump : ∀ {X : Object} (f : ℂ [ X , A ]) (g : ℂ [ X , B ])
%% → ∃![ f×g ] (ℂ [ fst ∘ f×g ] ≡ f P.× ℂ [ snd ∘ f×g ] ≡ g)
\end{align*}
$
$\pi$ is called the product (arrow) of $f$ and $g$.
\section{Monads}
%% \subsubsection{Functors}

35
doc/introduction.tex
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@ -73,12 +73,13 @@ the (left) identity law of the underlying category to proove $\idFun \comp g
%
\subsection{Equality of isomorphic types}
%
Let $\top$ denote the unit type -- a type with a single constructor. In the
propositions-as-types interpretation of type theory $\top$ is the proposition
that is always true. The type $A \x \top$ and $A$ has an element for each $a :
A$. So in a sense they are the same. The second element of the pair does not add
any ``interesting information''. It can be useful to identify such types. In
fact, it is quite commonplace in mathematics. Say we look at a set $\{x \mid
Let $\top$ denote the unit type -- a type with a single constructor. In
the propositions-as-types interpretation of type theory $\top$ is the
proposition that is always true. The type $A \x \top$ and $A$ has an element for
each $a : A$. So in a sense they are the same. The second element of the pair
does not add any ``interesting information''. It can be useful to identify such
types. In fact, it is quite commonplace in mathematics. Say we look at a set
$\{x \mid
\phi\ x \land \psi\ x\}$ and somehow conclude that $\psi\ x \equiv \top$ for all
$x$. A mathematician would immediately conclude $\{x \mid \phi\ x \land
\psi\ x\} \equiv \{x \mid \phi\ x\}$ without thinking twice. Unfortunately such
@ -92,8 +93,9 @@ types. The principle of univalence says that:
%
$$\mathit{univalence} \tp (A \simeq B) \simeq (A \equiv B)$$
%
In particular this allows us to construct an equality from an equivalence $\mathit{ua} \tp
(A \simeq B) \to (A \equiv B)$ and vice-versa.
In particular this allows us to construct an equality from an equivalence
($\mathit{ua} \tp (A \simeq B) \to (A \equiv B)$) and vice-versa.
\section{Formalizing Category Theory}
%
The above examples serve to illustrate the limitation of Agda. One case where
@ -115,20 +117,21 @@ Inspiration:
\end{verbatim}
The idea of formalizing Category Theory in proof assistants is not new. There
are a multitude of these available online. Just as first reference see this
question on Math Overflow: \cite{mo-formalizations}. Notably these two implementations of category theory in Agda:
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} -- setoid interpretation
\item
\url{https://github.com/pcapriotti/agda-categories} - homotopic setting with postulates
\url{https://github.com/pcapriotti/agda-categories} -- homotopic setting with postulates
\item
\url{https://github.com/pcapriotti/agda-categories} - homotopic setting in coq
\url{https://github.com/pcapriotti/agda-categories} -- homotopic setting in coq
\item
\url{https://github.com/mortberg/cubicaltt} - homotopic setting in \texttt{cubicaltt}
\url{https://github.com/mortberg/cubicaltt} -- homotopic setting in \texttt{cubicaltt}
\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.
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.
There are alternative approaches to working in a cubical setting where one can
still have univalence and functional extensionality. One option is to postulate

6
doc/macros.tex
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@ -59,3 +59,9 @@
\newcommand\rrr{\ggg}
\newcommand\fst{\mathit{fst}}
\newcommand\snd{\mathit{snd}}
\newcommand\Path{\mathit{Path}}
\newcommand\Category{\mathit{Category}}
\newcommand\TODO[1]{TODO: \emph{#1}}
\newcommand*{\QED}{\hfill\ensuremath{\square}}%
\newcommand\uexists{\exists!}
\newcommand\Arrow{\mathit{Arrow}}

4
doc/main.tex
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@ -26,8 +26,8 @@
\bibliographystyle{plainnat}
\nocite{cubical-demo}
\nocite{coquand-2013}
%% \bibliography{refs}
%% \begin{appendices}
\bibliography{refs}
\begin{appendices}
%% \input{planning.tex}
%% \input{halftime.tex}
\end{appendices}

40
src/Cat/Equivalence.agda
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@ -10,7 +10,7 @@ open import Cubical.GradLemma hiding (isoToPath)
open import Cat.Prelude using
( lemPropF ; setPi ; lemSig ; propSet
; Preorder ; equalityIsEquivalence ; propSig ; id-coe
; Setoid )
; Setoid ; _$_ ; propPi )
import Cubical.Univalence as U
@ -133,7 +133,7 @@ module _ {ℓa ℓb ℓ : Level} (A : Set ℓa) (B : Set ℓb) where
-- | The other inverse law does not hold in general, it does hold, however,
-- | if `A` and `B` are sets.
module _ (sA : isSet A) (sB : isSet B) where
module _ {f : A → B} (iso : Isomorphism f) where
module _ {f : A → B} where
module _ (iso-x iso-y : Isomorphism f) where
open Σ iso-x renaming (fst to x ; snd to inv-x)
open Σ iso-y renaming (fst to y ; snd to inv-y)
@ -146,22 +146,18 @@ module _ {ℓa ℓb ℓ : Level} (A : Set ℓa) (B : Set ℓb) where
y ∎
propInv : ∀ g → isProp (AreInverses f g)
propInv g t u i = a i , b i
propInv g t u = λ i → a i , b i
where
a : (fst t) ≡ (fst u)
a i = h
a i = funExt hh
where
hh : ∀ a → (g ∘ f) a ≡ a
hh a = sA ((g ∘ f) a) a (λ i → (fst t) i a) (λ i → (fst u) i a) i
h : g ∘ f ≡ idFun A
h i a = hh a i
b : (snd t) ≡ (snd u)
b i = h
b i = funExt hh
where
hh : ∀ b → (f ∘ g) b ≡ b
hh b = sB _ _ (λ i → snd t i b) (λ i → snd u i b) i
h : f ∘ g ≡ idFun B
h i b = hh b i
inx≡iny : (λ i → AreInverses f (fx≡fy i)) [ inv-x ≡ inv-y ]
inx≡iny = lemPropF propInv fx≡fy
@ -169,6 +165,7 @@ module _ {ℓa ℓb ℓ : Level} (A : Set ℓa) (B : Set ℓb) where
propIso : iso-x ≡ iso-y
propIso i = fx≡fy i , inx≡iny i
module _ (iso : Isomorphism f) where
inverse-to-from-iso : (toIso {f} ∘ fromIso {f}) iso ≡ iso
inverse-to-from-iso = begin
(toIso ∘ fromIso) iso ≡⟨⟩
@ -419,7 +416,7 @@ module _ {ℓa ℓb : Level} {A : Set ℓa} {P : A → Set ℓb} where
equivPropSig pA p q = fromIsomorphism _ _ iso
where
f : ∀ {p q} → p ≡ q → fst p ≡ fst q
f e i = fst (e i)
f = cong fst
g : ∀ {p q} → fst p ≡ fst q → p ≡ q
g {p} {q} = lemSig pA p q
ve-re : (e : p ≡ q) → (g ∘ f) e ≡ e
@ -507,31 +504,26 @@ module _ {ℓa ℓb : Level} {A : Set ℓa} {P : A → Set ℓb} where
→ ((a : A) → P a ≃ Q a) → Σ A P ≃ Σ A Q
equivSig {Q = Q} eA = res
where
P≅Q : ∀ {a} → P a ≅ Q a
P≅Q {a} = toIsomorphism _ _ (eA a)
f : Σ A P → Σ A Q
f (a , pA) = a , fst (eA a) pA
f (a , pA) = a , fst P≅Q pA
g : Σ A Q → Σ A P
g (a , qA) = a , g' qA
where
k : Isomorphism _
k = toIso _ _ (snd (eA a))
open Σ k renaming (fst to g')
g (a , qA) = a , fst (snd P≅Q) qA
ve-re : (x : Σ A P) → (g ∘ f) x ≡ x
ve-re x i = fst x , eq i
ve-re (a , pA) i = a , eq i
where
eq : snd ((g ∘ f) x) ≡ snd x
eq : snd ((g ∘ f) (a , pA)) ≡ pA
eq = begin
snd ((g ∘ f) x) ≡⟨⟩
snd ((g ∘ f) (a , pA)) ≡⟨⟩
snd (g (f (a , pA))) ≡⟨⟩
g' (fst (eA a) pA) ≡⟨ lem ⟩
pA ∎
where
open Σ x renaming (fst to a ; snd to pA)
k : Isomorphism _
k = toIso _ _ (snd (eA a))
open Σ k renaming (fst to g' ; snd to inv)
open Σ (snd P≅Q) renaming (fst to g' ; snd to inv)
-- anti-funExt
lem : (g' ∘ (fst (eA a))) pA ≡ pA
lem i = fst inv i pA
lem = cong (_$ pA) (fst (snd (snd P≅Q)))
re-ve : (x : Σ A Q) → (f ∘ g) x ≡ x
re-ve x i = fst x , eq i
where

4
src/Cat/Prelude.agda
View file

@ -62,10 +62,8 @@ module _ (ℓ : Level) where
syntax ∃!-syntax (λ x → B) = ∃![ x ] B
module _ {ℓa ℓb} {A : Set ℓa} {P : A → Set ℓb} (f g : ∃! P) where
open Σ (snd f) renaming (snd to u)
∃-unique : fst f ≡ fst g
∃-unique = u (fst (snd g))
∃-unique = (snd (snd f)) (fst (snd g))
module _ {ℓa ℓb : Level} {A : Set ℓa} {B : A → Set ℓb} {a b : Σ A B}
(fst≡ : (λ _ → A) [ fst a ≡ fst b ])