Stuff about product-category
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% Requires: hypperref
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\newcommand*{\institution}[1]{\gdef\@institution{#1}}
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\renewcommand*{\maketitle}{%
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\begin{titlepage}
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% IMPRINT PAGE (BACK OF TITLE PAGE)
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\newpage
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\thispagestyle{plain}
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\vspace*{4.5cm}
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\@title\\
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\textit{\@title}\\
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\@subtitle\\
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\copyright\ \the\year ~ \MakeUppercase{\@author}
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\vspace{4.5cm}
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\setlength{\parskip}{0.5cm}
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\textbf{Author:}\\
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\@author\\
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\href{mailto:\@authoremail>}{\texttt{<\@authoremail>}}
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\copyright ~ \MakeUppercase{\@author}, \the\year.
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\textbf{Supervisor:}\\
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\@supervisor\\
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\href{mailto:\@supervisoremail>}{\texttt{<\@supervisoremail>}}\\
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@ -67,20 +90,18 @@
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\textbf{Examiner:}\\
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\@examiner\\
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\href{mailto:\@examineremail>}{\texttt{<\@examineremail>}}\\
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\@examinerdepartment\setlength{\parskip}{1cm}
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\@examinerdepartment
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\vfill
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Master's Thesis \the\year\\ % Report number currently not in use
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\@department\\
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%Division of Division name\\
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%Name of research group (if applicable)\\
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SE-412 96 Gothenburg\\
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Telephone +46 31 772 1000 \setlength{\parskip}{0.5cm}
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\vfill
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Telephone +46 31 772 1000 \setlength{\parskip}{0.5cm}\\
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% Caption for cover page figure if used, possibly with reference to further information in the report
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%% Cover: Wind visualization constructed in Matlab showing a surface of constant wind speed along with streamlines of the flow. \setlength{\parskip}{0.5cm}
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%Printed by [Name of printing company]\\
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Gothenburg, Sweden \the\year
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@ -66,7 +66,7 @@ $$
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$$
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%
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The two types are logically equivalent, however. One can construct the latter
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from the formerr simply by ``forgetting'' that $\idToIso$ plays the role
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from the former simply by ``forgetting'' that $\idToIso$ plays the role
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of the equivalence. The other direction is more involved.
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With all this in place it is now possible to prove that all the laws are indeed
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@ -367,9 +367,10 @@ $$
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$$
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%
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%
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$$
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\begin{align}
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\label{eq:coeCod}
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\mathit{coeCod} \tp \prod_{f \tp A \to X} \mathit{coe}\ p_{\mathit{cod}}\ f \equiv \iota \lll f
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$$
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\end{align}
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%
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I will give the proof of the first theorem here, the second one is analagous.
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\begin{align*}
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@ -452,7 +453,8 @@ arguments. Or in other words since $\Hom_{A\ B}$ is a set for all $A\ B \tp
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\Object$ then so is $\Hom_{B\ A}$.
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Now, to show that this category is univalent is not as straight-forward. Lucliy
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section \ref{sec:equiv} gave us some tools to work with equivalences. We saw that we
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section \ref{sec:equiv} gave us some tools to work with equivalences.
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We saw that we
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can prove this category univalent by giving an inverse to
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$\idToIso_{\mathit{Op}} \tp (A \equiv B) \to (A \approxeq_{\mathit{Op}} B)$.
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From the original category we have that $\idToIso \tp (A \equiv B) \to (A \cong
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@ -692,9 +694,10 @@ then we know that products also are propositions. But before we get to that,
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let's recall the definition of products.
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\subsection{Products}
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Given a category $\bC$ and two objects $A$ and $B$ in $bC$ we define the product
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of $A$ and $B$ to be an object $A \x B$ in $\bC$ and two arrows $\pi_1 \tp A \x
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B \to A$ and $\pi_2 \tp A \x B \to B$ called the projections of the product. The projections must satisfy the following property:
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Given a category $\bC$ and two objects $A$ and $B$ in $\bC$ we define the
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product of $A$ and $B$ to be an object $A \x B$ in $\bC$ and two arrows $\pi_1
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\tp A \x B \to A$ and $\pi_2 \tp A \x B \to B$ called the projections of the
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product. The projections must satisfy the following property:
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For all $X \tp Object$, $f \tp \Arrow\ X\ A$ and $g \tp \Arrow\ X\ B$ we have
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that there exists a unique arrow $\pi \tp \Arrow\ X\ (A \x B)$ satisfying
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@ -706,8 +709,8 @@ that there exists a unique arrow $\pi \tp \Arrow\ X\ (A \x B)$ satisfying
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\pi_1 \lll \pi \equiv f \x \pi_2 \lll \pi \equiv g
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%% ump : ∀ {X : Object} (f : ℂ [ X , A ]) (g : ℂ [ X , B ])
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%% → ∃![ f×g ] (ℂ [ fst ∘ f×g ] ≡ f P.× ℂ [ snd ∘ f×g ] ≡ g)
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\end{align*}
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$
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\end{align}
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%
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$\pi$ is called the product (arrow) of $f$ and $g$.
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\subsection{Pair category}
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@ -715,7 +718,7 @@ $\pi$ is called the product (arrow) of $f$ and $g$.
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\newcommand\pairA{\mathcal{A}}
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\newcommand\pairB{\mathcal{B}}
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Given a base category $\bC$ and two objects in this category $\pairA$ and
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$\pairrB$ we can construct the ``pair category'': \TODO{This is a working title,
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$\pairB$ we can construct the ``pair category'': \TODO{This is a working title,
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it's nice to have a name for this thing to refer back to}
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The type objects in this category will be an object in the underlying category,
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@ -731,11 +734,9 @@ category will consist of an arrow from the underlying category $\pairf \tp
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\Arrow\ A\ B$ satisfying:
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%
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\begin{align}
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\begin{split}
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\label{eq:pairArrowLaw}
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b_0 \lll f & \equiv a_0 \\
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b_1 \lll f & \equiv a_1
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\end{split}
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b_0 \lll f \equiv a_0 \x
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b_1 \lll f \equiv a_1
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\end{align}
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The identity morphism is the identity morphism from the underlying category.
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@ -758,6 +759,160 @@ choose $g \lll f$ and we must now verify that it satisfies
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a_0
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&& \text{$g$ satisfies \ref{eq:pairArrowLaw}} \\
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\end{align*}
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%
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Now we must verify the category-laws. For all the laws we will follow the
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pattern of using the law from the underlying category, and that the type of
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arrows form a set. For instance, to prove associativity we must prove that
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%
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\begin{align}
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\label{eq:productAssoc}
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\overline{h} \lll (\overline{g} \lll \overline{f})
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\equiv
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(\overline{h} \lll \overline{g}) \lll \overline{f}
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\end{align}
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%
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Herer $\lll$ refers to the `embellished' composition abd $\overline{f}$,
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$\overline{g}$ and $\overline{h}$ are triples consisting of arrows from the
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underlying category ($f$, $g$ and $h$) and a pair of witnesses to
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\ref{eq:pairArrowLaw}.
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%% Luckily those winesses are paths in the hom-set of the
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%% underlying category which is a set, so these are mere propositions.
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The proof
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obligations is:
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%
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\begin{align}
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\label{eq:productAssocUnderlying}
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h \lll (g \lll f)
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\equiv
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(h \lll g) \lll f
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\end{align}
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%
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Which is provable by and that the witness to \ref{eq:pairArrowLaw} for the
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left-hand-side and the right-hand-side are the same. The type of this goal is
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quite involved, and I will not write it out in full, but it suffices to show the
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type of the path-space. Note that the arrows in \ref{eq:productAssoc} are arrows
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from $\mathcal{A} = (A , a_\pairA , a_\pairB)$ to $\mathcal{D} = (D , d_\pairA ,
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d_\pairB)$ where $a_\pairA$, $a_\pairB$, $d_\pairA$ and $d_\pairB$ are arrows in
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the underlying category. Given that $p$ is the proof of
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\ref{eq:productAssocUnderlying} we then have that the witness to
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\ref{eq:pairArrowLaw} vary over the type:
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%
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\begin{align}
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\label{eq:productPath}
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λ\ i → d_\pairA \lll p\ i ≡ a_\pairA × d_\pairB \lll p\ i ≡ a_\pairB
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\end{align}
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%
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And these paths are in the type of the hom-set of the underlying category, so
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they are mere propositions. We cannot apply the fact that arrows in $\bC$ are
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sets directly, however, since $\isSet$ only talks about non-dependent paths, in
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stead we generalize \ref{eq:productPath} to:
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%
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\begin{align}
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\label{eq:productEqPrinc}
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\prod_{f \tp \Arrow\ X\ Y} \isProp\ \left( y_\pairA \lll f ≡ x_\pairA × y_\pairB \lll f ≡ x_\pairB \right)
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\end{align}
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%
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For all objects $X , x_\pairA , x_\pairB$ and $Y , y_\pairA , y_\pairB$, but
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this follows from pairs preserving homotopical structure and arrows in the
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underlying category being sets. This gives us an equality principle for arrows
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in this category that says that to prove two arrows $f, f_0, f_1$ and $g, g_0,
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$g_1$ equal it suffices to give a proof that $f$ and $g$ are equal.
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%% %
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%% $$
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%% \prod_{(f, f_0, f_1)\; (g,g_0,g_1) \tp \Arrow\ X\ Y} f \equiv g \to (f, f_0, f_1) \equiv (g,g_0,g_1)
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%% $$
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%% %
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And thus we have proven \ref{eq:productAssoc} simply with
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\ref{eq:productAssocUnderlying}.
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Now we must prove that arrows form a set:
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%
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$$
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\isSet\ (\Arrow\ \mathcal{X}\ \mathcal{Y})
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$$
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%
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Since pairs preserve homotopical structure this reduces to:
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%
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$$
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\isSet\ (\Arrow_\bC\ X\ Y)
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$$
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%
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Which holds. And
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%
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$$
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\prod_{f \tp \Arrow\ X\ Y} \isSet\ \left( y_\pairA \lll f ≡ x_\pairA × y_\pairB \lll f ≡ x_\pairB \right)
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$$
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%
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This we get from \ref{eq:productEqPrinc} and the fact that homotopical structure
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is cumulative.
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This finishes the proof that this is a valid pre-category.
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\subsubsection{Univalence}
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To prove that this is a proper category it must be shown that it is univalent.
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That is, for any two objects $\mathcal{X} = (X, x_\mathcal{A} , x_\mathca{B})$
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and $\mathcal{Y} = Y, y_\mathcal{A}, y_\mathcal{B}$ I will show:
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%
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\begin{align}
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(\mathcal{X} \equiv \mathcal{Y}) \cong (\mathcal{X} \approxeq \mathcal{Y})
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\end{align}
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I do this by showing that the following sequence of types are isomorphic.
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The first type is:
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%
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\begin{align}
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\label{eq:univ-0}
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(X , x_\mathcal{A} , x_\mathcal{B}) ≡ (Y , y_\mathcal{A} , y_\mathcal{B})
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\end{align}
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%
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The next types will be the triple:
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%
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\begin{align}
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\label{eq:univ-1}
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\begin{split}
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p \tp & X \equiv Y \\
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& \Path\ (λ i → \Arrow\ (p\ i)\ \mathcal{A})\ x_\mathcal{A}\ y_\mathcal{A} \\
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& \Path\ (λ i → \Arrow\ (p\ i)\ \mathcal{B})\ x_\mathcal{B}\ y_\mathcal{B}
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\end{split}
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%% \end{split}
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\end{align}
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The next type is very similar, but in stead of a path we will have an
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isomorphism, and create a path from this:
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%
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\begin{align}
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\label{eq:univ-2}
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\begin{split}
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\var{iso} \tp & X \cong Y \\
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& \Path\ (λ i → \Arrow\ (\widetilde{p}\ i)\ \mathcal{A})\ x_\mathcal{A}\ y_\mathcal{A} \\
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& \Path\ (λ i → \Arrow\ (\widetilde{p}\ i)\ \mathcal{B})\ x_\mathcal{B}\ y_\mathcal{B}
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\end{split}
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\end{align}
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%
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Where $\widetilde{p} \defeq \var{isoToId}\ \var{iso} \tp X \equiv Y$.
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Finally we have the type:
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%
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\begin{align}
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\label{eq:univ-3}
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(X , x_\mathcal{A} , x_\mathcal{B}) ≊ (Y , y_\mathcal{A} , y_\mathcal{B})
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\end{align}
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\emph{Proposition} \ref{eq:univ-0} is isomorphic to \ref{eq:univ-1}: This is
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just an application of the fact that a path between two pairs $a_0, a_1$ and
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$b_0, b_1$ corresponds to a pair of paths between $a_0,b_0$ and $a_1,b_1$ (check
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the implementation for the details).
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\emph{Proposition} \ref{eq:univ-1} is isomorphic to \ref{eq:univ-2}:
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\TODO{Super complicated}
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\emph{Proposition} \ref{eq:univ-2} is isomorphic to \ref{eq:univ-3}:
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For this I will two corrolaries of \ref{eq:coeCod}:
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%
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\begin{align}
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\label{domain-twist}
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\end{align}
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\section{Monads}
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@ -30,7 +30,7 @@ which is:
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%
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\newcommand{\suc}[1]{\mathit{suc}\ #1}
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\begin{align*}
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+ & : \bN \to \bN \\
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+ & : \bN \to \bN \to \bN \\
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n + 0 & \defeq n \\
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n + (\suc{m}) & \defeq \suc{(n + m)}
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\end{align*}
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@ -69,7 +69,7 @@ The proof obligation that this satisfies the identity law of functors
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%
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One needs functional extensionality to ``go under'' the function arrow and apply
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the (left) identity law of the underlying category to proove $\idFun \comp g
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\equiv g$ and thus closing the.
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\equiv g$ and thus closing the goal.
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%
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\subsection{Equality of isomorphic types}
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%
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@ -3,9 +3,9 @@
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\newcommand{\coloneqq}{\mathrel{\vcenter{\baselineskip0.5ex \lineskiplimit0pt
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\hbox{\scriptsize.}\hbox{\scriptsize.}}}%
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=}
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=}
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\newcommand{\defeq}{\coloneqq}
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\newcommand{\defeq}{\triangleq}
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\newcommand{\bN}{\mathbb{N}}
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\newcommand{\bC}{\mathbb{C}}
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\newcommand{\bX}{\mathbb{X}}
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24
doc/main.tex
24
doc/main.tex
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@ -4,9 +4,29 @@
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\input{packages.tex}
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\input{macros.tex}
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\title{Univalent categories in cubical Agda}
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\subtitle{}
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\title{Univalent Categories in Cubical Agda}
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\author{Frederik Hanghøj Iversen}
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%% \usepackage[
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%% subtitle=foo,
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%% author=Frederik Hanghøj Iversen,
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%% authoremail=hanghj@student.chalmers.se,
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%% newcommand=chalmers,=Chalmers University of Technology,
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%% supervisor=Thierry Coquand,
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%% supervisoremail=coquand@chalmers.se,
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%% supervisordepartment=chalmers,
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%% cosupervisor=Andrea Vezzosi,
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%% cosupervisoremail=vezzosi@chalmers.se,
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%% cosupervisordepartment=chalmers,
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%% examiner=Andreas Abel,
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%% examineremail=abela@chalmers.se,
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%% examinerdepartment=chalmers,
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%% institution=chalmers,
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%% department=Department of Computer Science and Engineering,
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%% researchgroup=Programming Logic Group
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%% ]{chalmerstitle}
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\usepackage{chalmerstitle}
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\subtitle{}
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\authoremail{hanghj@student.chalmers.se}
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\newcommand{\chalmers}{Chalmers University of Technology}
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\supervisor{Thierry Coquand}
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@ -28,15 +28,17 @@
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\usepackage{listings}
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\usepackage{fancyvrb}
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\usepackage{chalmerstitle}
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\usepackage{mathpazo}
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\usepackage[scaled=0.95]{helvet}
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\usepackage{courier}
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\linespread{1.05} % Palatino looks better with this
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\usepackage{lmodern}
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\usepackage{fontspec}
|
||||
\setmonofont[Mapping=tex-text]{FreeMono.otf}
|
||||
\usepackage{sourcecodepro}
|
||||
%% \setmonofont{Latin Modern Mono}
|
||||
%% \setmonofont[Mapping=tex-text]{FreeMono.otf}
|
||||
%% \setmonofont{FreeMono.otf}
|
||||
|
||||
|
||||
|
|
@ -44,3 +46,9 @@
|
|||
\setlength{\headheight}{15pt}
|
||||
\renewcommand{\chaptermark}[1]{\markboth{\textsc{Chapter \thechapter. #1}}{}}
|
||||
\renewcommand{\sectionmark}[1]{\markright{\textsc{\thesection\ #1}}}
|
||||
|
||||
% Allows for the use of unicode-letters:
|
||||
\usepackage{unicode-math}
|
||||
|
||||
|
||||
%% \RequirePackage{kvoptions}
|
||||
|
|
|
|||
Loading…
Reference in a new issue