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@ -1,4 +1,5 @@
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This implementation formalizes the following concepts:
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\chapter{Implementation}
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This implementation formalizes the following concepts $>\!\!>\!\!=$:
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%
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\begin{itemize}
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\item Core categorical concepts
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@ -224,11 +225,18 @@ $$
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$$
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%
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One application of this, and a perhaps somewhat surprising result, is that
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terminal objects are propositional. Intuitively; they do not have any
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interesting structure. The proof of this follows from the usual observation that
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any two terminal objects are isomorphic. The proof is omitted here, but the
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curious reader can check the implementation for the details. \TODO{The proof is
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a bit fun, should I include it?}
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terminal objects are propositional. Intuitively; they do not
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have any interesting structure:
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%
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\begin{align}
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\label{eq:termProp}
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\isProp\ \var{Terminal}
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\end{align}
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%
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The proof of this follows from the usual
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observation that any two terminal objects are isomorphic. The proof is omitted
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here, but the curious reader can check the implementation for the details.
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\TODO{The proof is a bit fun, should I include it?}
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\section{Equivalences}
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\label{sec:equiv}
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@ -660,9 +668,10 @@ inverse to $f$. $p$ is inhabited by:
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For the other (dependent) path we can prove that being-an-inverse-of is a
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proposition and then use $\lemPropF$. So we prove the generalization:
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%
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\begin{align*}
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\begin{align}
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\label{eq:propAreInversesGen}
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\prod_{g : B \to A} \isProp\ (\mathit{AreInverses}\ f\ g)
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\end{align*}
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\end{align}
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%
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But $\mathit{AreInverses}\ f\ g$ is a pair of equations on arrows, so we use
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$\propSig$ and the fact that both $A$ and $B$ are sets to close this proof.
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@ -689,7 +698,7 @@ $$
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Where $\mathit{Product}\ \bC\ A\ B$ denotes the type of products of objects $A$
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and $B$ in the category $\bC$. I do this by constructing a category whose
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terminal objects are equivalent to products in $\bC$, and since terminal objects
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are propositional in a proper category and equivalences preservehomotopy level,
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are propositional in a proper category and equivalences preserve homotopy level,
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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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@ -721,8 +730,8 @@ Given a base category $\bC$ and two objects in this category $\pairA$ and
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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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$X$, and two arrows (also from the underlying category)
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The type of objects in this category will be an object in the underlying
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category, $X$, and two arrows (also from the underlying category)
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$\Arrow\ X\ \pairA$ and $\Arrow\ X\ \pairB$.
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\newcommand\pairf{\ensuremath{f}}
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@ -771,7 +780,7 @@ arrows form a set. For instance, to prove associativity we must prove that
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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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Herer $\lll$ refers to the `embellished' composition and $\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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@ -787,15 +796,15 @@ h \lll (g \lll f)
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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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Which is provable by \TODO{What?} and that the witness to \ref{eq:pairArrowLaw}
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for the left-hand-side and the right-hand-side are the same. The type of this
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goal is quite involved, and I will not write it out in full, but at present it
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suffices to show the type of the path-space. Note that the arrows in
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\ref{eq:productAssoc} are arrows from $\mathcal{A} = (A , a_\pairA , a_\pairB)$
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to $\mathcal{D} = (D , d_\pairA , d_\pairB)$ where $a_\pairA$, $a_\pairB$,
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$d_\pairA$ and $d_\pairB$ are arrows in the underlying category. Given that $p$
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is the chosen proof of \ref{eq:productAssocUnderlying} we then have that the
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witness to \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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@ -907,156 +916,318 @@ 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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\emph{Proposition} \ref{eq:univ-2} is isomorphic to \ref{eq:univ-3}: For this I
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will swho two corrolaries of \ref{eq:coeCod}: For an isomorphism $(\iota,
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\inv{\iota}, \var{inv}) \tp A \cong B$, arrows $f \tp \Arrow\ A\ X$, $g \tp
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\Arrow\ B\ X$ and a heterogeneous path between them, $q \tp \Path\ (\lambda i
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\to p_\var{dom}\ i)\ f\ g$, where $p_\var{dom} \tp \Arrow\ A\ X \equiv
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\Arrow\ B\ X$ is a path induced by $\var{iso}$, we have the following two
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results
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%
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\begin{align}
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\label{domain-twist}
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\label{eq:domain-twist-0}
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f & \equiv g \lll \iota \\
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\label{eq:domain-twist-1}
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g & \equiv f \lll \inv{\iota}
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\end{align}
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%
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Proof: \TODO{\ldots}
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Now we can prove the equiavalence in the following way: Given $(f, \inv{f},
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\var{inv}_f) \tp X \cong Y$ and two heterogeneous paths
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%
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\begin{align*}
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p_\mathcal{A} & \tp \Path\ (\lambda i \to p_\var{dom}\ i)\ x_\mathcal{A}\ y_\mathcal{A}\\
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%
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q_\mathcal{B} & \tp \Path\ (\lambda i \to p_\var{dom}\ i)\ x_\mathcal{B}\ y_\mathcal{B}
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\end{align*}
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%
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all as in \ref{eq:univ-2}. I use $p_\var{dom}$ here again to mean the path
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induced by the isomorphism $f, \inv{f}$. I must now construct an isomorphism
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$(X, x_\mathcal{A}, x_\mathcal{B}) \approxeq (Y, y_\mathcal{A}, y_\mathcal{B}$
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as in \ref{eq:univ-3}. That is, an isomorphism in the present category. I remind
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the reader that such a gadget is a triple. The first component shall be:
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%
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\begin{align}
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f \tp \Arrow\ X\ Y
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\end{align}
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%
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To show that this choice fits the bill I must now verify that it satisfies
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\ref{eq:pairArrowLaw}, which in this case becomes:
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%
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\begin{align}
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y_\mathcal{A} \lll f ≡ x_\mathcal{A} × y_\mathcal{B} \lll f ≡ x_\mathcal{B}
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\end{align}
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%
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Which, since $f$ is an isomorphism and $p_\mathcal{A}$ (resp. $p_\mathcal{B}$)
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is a path varying according to a path constructed from this isomorphism, this is
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exactly what \ref{eq:domain-twist-0} gives us.
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%
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The other direction is quite analagous. We choose $\inv{f}$ as the morphism and
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prove that it satisfies \ref{eq:pairArrowLaw} with \ref{eq:domain-twist-1}.
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We must now show that this choice of arrows indeed form an isomorphism. Our
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equality principle for arrows in this category (\ref{eq:productEqPrinc}) gives
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us that it suffices to show that $f$ and $\inv{f}$, this is exactly
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$\var{inv}_f$.
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This concludes the first direction of the isomorphism that we're constructing.
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For the other direction we're given just given the isomorphism
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%
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$$
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(f, \inv{f}, \var{inv}_f)
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\tp
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(X, x_\mathcal{A}, x_\mathcal{B}) \approxeq (Y, y_\mathcal{A}, y_\mathcal{B})
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$$
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%
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Projecting out the first component gives us the isomorphism
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%
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$$
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(\fst\ f, \fst\ \inv{f}, \congruence\ \fst\ \var{inv}_f, \congruence\ \fst\ \var{inv}_{\inv{f}})
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\tp X \approxeq Y
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$$
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%
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This gives rise to the following paths:
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%
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\begin{align}
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\begin{split}
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\widetilde{p} & \tp X \equiv Y \\
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\widetilde{p}_\mathcal{A} & \tp \Arrow\ X\ \mathcal{A} \equiv \Arrow\ Y\ \mathcal{A} \\
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\widetilde{p}_\mathcal{B} & \tp \Arrow\ X\ \mathcal{B} \equiv \Arrow\ Y\ \mathcal{B}
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\end{split}
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\end{align}
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%
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It then remains to construct the two paths:
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%
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\begin{align}
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\begin{split}
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\label{eq:product-paths}
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& \Path\ (λ i → \widetilde{p}_\mathcal{A}\ i)\ x_\mathcal{A}\ y_\mathcal{A}\\
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& \Path\ (λ i → \widetilde{p}_\mathcal{B}\ i)\ 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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This is achieved with the following lemma:
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%
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\begin{align}
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\prod_{a \tp A} \prod_{b \tp B} \prod_{q \tp A \equiv B} \var{coe}\ q\ a ≡ b →
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\Path\ (λ i → q\ i)\ a\ b
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\end{align}
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%
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Which is used without proof. See the implementation for the details.
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\ref{eq:product-paths} is the proven with the propositions:
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%
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\begin{align}
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\begin{split}
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\label{eq:product-paths}
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\var{coe}\ \widetilde{p}_\mathcal{A}\ x_\mathcal{A} ≡ y_\mathcal{A}\\
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\var{coe}\ \widetilde{p}_\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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The proof of the first one is:
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%
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\begin{align*}
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\var{coe}\ \widetilde{p}_\mathcal{A}\ x_\mathcal{A}
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& ≡ x_\mathcal{A} \lll \fst\ \inv{f} && \text{$\mathit{coeDom}$ and the isomorphism $f, \inv{f}$} \\
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& ≡ y_\mathcal{A} && \text{\ref{eq:pairArrowLaw} for $\inv{f}$}
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\end{align*}
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%
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We have now constructed the maps between \ref{eq:univ-0} and \ref{eq:univ-1}. It
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remains to show that they are inverses of each other. To cut a long story short,
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the proof uses the fact that isomorphism-of is propositional and that arrows (in
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both categories) are sets. The reader is refered to the implementation for the
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gory details.
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%
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\subsection{Propositionality of products}
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%
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Now that we've constructed the ``pair category'' I'll demonstrate how to use
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this to prove that products are propositional. I will do this by showing that
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terminal objects in this category are equivalent to products:
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%
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\begin{align}
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\var{Terminal} ≃ \var{Product}\ ℂ\ \mathcal{A}\ \mathcal{B}
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\end{align}
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%
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And as always we do this by constructing an isomorphism:
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%
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In the direction $\var{Terminal} → \var{Product}\ ℂ\ \mathcal{A}\ \mathcal{B}$
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we're given a terminal object $X, x_𝒜, x_ℬ$. $X$ Will be the product-object and
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$x_𝒜, x_ℬ$ will be the product arrows, so it just remains to verify that this is
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indeed a product. That is, for an object $Y$ and two arrows $y_𝒜 \tp
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\Arrow\ Y\ 𝒜$, $y_ℬ\ \Arrow\ Y\ ℬ$ we must find a unique arrow $f \tp
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\Arrow\ Y\ X$ satisfying:
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%
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\begin{align}
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\label{eq:pairCondRev}
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\begin{split}
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x_𝒜 \lll f & ≡ y_𝒜 \\
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x_ℬ \lll f & ≡ y_ℬ
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\end{split}
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\end{align}
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%
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Since $X, x_𝒜, x_ℬ$ is a terminal object there is a \emph{unique} arrow from
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this object to any other object, so also $Y, y_𝒜, y_ℬ$ in particular (which is
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also an object in the pair category). The arrow we will play the role of $f$ and
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it immediately satisfies \ref{eq:pairCondRev}. Any other arrow satisfying these
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conditions will be equal since $f$ is unique.
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For the other direction we are now given a product $X, x_𝒜, x_ℬ$. Again this
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will be the terminal object. So now it remains that for any other object there
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is aunique arrow from that object into $X, x_𝒜, x_ℬ$. Let $Y, y_𝒜, y_ℬ$ be
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another object. As the arrow $\Arrow\ Y\ X$ we choose the product-arrow $y_𝒜 \x
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y_ℬ$. Since this is a product-arrow it satisfies \ref{eq:pairCondRev}. Let us
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name the witness to this $\phi_{y_𝒜 \x y_ℬ}$. So we have picked as our center of
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contraction $y_𝒜 \x y_ℬ , \phi_{y_𝒜 \x y_ℬ}$ we must now show that it is
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contractible. So let $f \tp \Arrow\ X\ Y$ and $\phi_f$ be given (here $\phi_f$
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is the proof that $f$ satisfies \ref{eq:pairCondRev}). The proof will be a pair
|
|
|
|
|
of proofs:
|
|
|
|
|
%
|
|
|
|
|
\begin{alignat}{3}
|
|
|
|
|
p \tp & \Path\ (\lambda i \to \Arrow\ X\ Y)\quad
|
|
|
|
|
&& f\quad && y_𝒜 \x y_ℬ \\
|
|
|
|
|
& \Path\ (\lambda i \to \Phi\ (p\ i))\quad
|
|
|
|
|
&& \phi_f\quad && \phi_{y_𝒜 \x y_ℬ}
|
|
|
|
|
\end{alignat}
|
|
|
|
|
%
|
|
|
|
|
Here $\Phi$ is given as:
|
|
|
|
|
$$
|
|
|
|
|
\prod_{f \tp \Arrow\ Y\ X}
|
|
|
|
|
x_𝒜 \lll f ≡ y_𝒜
|
|
|
|
|
× x_ℬ \lll f ≡ y_ℬ
|
|
|
|
|
$$
|
|
|
|
|
%
|
|
|
|
|
$p$ follows from the universal property of $y_𝒜 \x y_ℬ$. For the latter we will
|
|
|
|
|
again use the same trick we did in \ref{eq:propAreInversesGen} and prove this
|
|
|
|
|
more general result:
|
|
|
|
|
%
|
|
|
|
|
$$
|
|
|
|
|
\prod_{f \tp \Arrow\ Y\ X} \isProp\ (
|
|
|
|
|
x_𝒜 \lll f ≡ y_𝒜
|
|
|
|
|
× x_ℬ \lll f ≡ y_ℬ
|
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|
|
|
)
|
|
|
|
|
$$
|
|
|
|
|
%
|
|
|
|
|
Which follows from arrows being sets and pairs preserving such. Thus we can
|
|
|
|
|
close the final proof with an application of $\lemPropF$.
|
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|
|
|
|
|
|
|
|
This concludes the proof $\var{Terminal} ≃
|
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|
|
|
\var{Product}\ ℂ\ \mathcal{A}\ \mathcal{B}$ and since we have that equivalences
|
|
|
|
|
preserve homotopic levels along with \ref{eq:termProp} we get our final result.
|
|
|
|
|
That in any category:
|
|
|
|
|
%
|
|
|
|
|
\begin{align}
|
|
|
|
|
\prod_{A\ B \tp \Object} \isProp\ (\var{Product}\ \bC\ A\ B)
|
|
|
|
|
\end{align}
|
|
|
|
|
%
|
|
|
|
|
\section{Monads}
|
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|
|
|
In this section I show two formulations of monads and then show that they are
|
|
|
|
|
the same. The two representations are referred to as the monoidal- and kleisli-
|
|
|
|
|
representation respectively.
|
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|
|
|
|
|
|
|
%% \subsubsection{Functors}
|
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|
|
|
%% Defines the notion of a functor - also split up into data and laws.
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|
|
We shall let a category $\bC$ be given. In the remainder all objects and arrows
|
|
|
|
|
will implicitly refer to objects and arrows in this category.
|
|
|
|
|
%
|
|
|
|
|
\subsection{Monoidal formulation}
|
|
|
|
|
The monoidal formulation of monads consists of the following data:
|
|
|
|
|
%
|
|
|
|
|
\begin{align}
|
|
|
|
|
\label{eq:monad-monoidal-data}
|
|
|
|
|
\begin{split}
|
|
|
|
|
\EndoR & \tp \Endo ℂ \\
|
|
|
|
|
\var{pure} & \tp \NT{\EndoR^0}{\EndoR} \\
|
|
|
|
|
\var{join} & \tp \NT{\EndoR^2}{\EndoR}
|
|
|
|
|
\end{split}
|
|
|
|
|
\end{align}
|
|
|
|
|
%
|
|
|
|
|
Here $\NTsym$ denotes natural transformations, the super-script in $\EndoR^2$
|
|
|
|
|
Denotes the composition of $\EndoR$ with itself. By the same token $\EndoR^0$ is
|
|
|
|
|
a curious way of denoting the identity functor. This notation has been chosen
|
|
|
|
|
for didactic purposes.
|
|
|
|
|
|
|
|
|
|
%% Propositionality for being a functor.
|
|
|
|
|
Denote the arrow-map of $\EndoR$ as $\fmap$, then this data must satisfy the
|
|
|
|
|
following laws:
|
|
|
|
|
%
|
|
|
|
|
\begin{align}
|
|
|
|
|
\label{eq:monad-monoidal-laws}
|
|
|
|
|
\begin{split}
|
|
|
|
|
\var{join} \lll \fmap\ \var{join}
|
|
|
|
|
& ≡ \var{join} \lll \var{join}\ \fmap \\
|
|
|
|
|
\var{join} \lll \var{pure}\ \fmap & ≡ \identity \\
|
|
|
|
|
\var{join} \lll \fmap\ \var{pure} & ≡ \identity
|
|
|
|
|
\end{split}
|
|
|
|
|
\end{align}
|
|
|
|
|
%
|
|
|
|
|
The implicit arguments to the arrows above have been left out and the objects
|
|
|
|
|
they range over are universally quantified.
|
|
|
|
|
|
|
|
|
|
%% Composition of functors and the identity functor.
|
|
|
|
|
\subsection{Kleisli formulation}
|
|
|
|
|
%
|
|
|
|
|
The kleisli-formulation consists of the following data:
|
|
|
|
|
%
|
|
|
|
|
\begin{align}
|
|
|
|
|
\label{eq:monad-kleisli-data}
|
|
|
|
|
\begin{split}
|
|
|
|
|
\EndoR & \tp \Object → \Object \\
|
|
|
|
|
\pure & \tp % \prod_{X \tp Object}
|
|
|
|
|
\Arrow\ X\ (\EndoR\ X) \\
|
|
|
|
|
\bind & \tp % \prod_{X\;Y \tp Object} → \Arrow\ X\ (\EndoR\ Y)
|
|
|
|
|
\Arrow\ (\EndoR\ X)\ (\EndoR\ Y)
|
|
|
|
|
\end{split}
|
|
|
|
|
\end{align}
|
|
|
|
|
%
|
|
|
|
|
The objects $X$ and $Y$ are implicitly universally quantified.
|
|
|
|
|
|
|
|
|
|
%% \subsubsection{Products}
|
|
|
|
|
%% Definition of what it means for an object to be a product in a given category.
|
|
|
|
|
It's interesting to note here that this formulation does not talk about natural
|
|
|
|
|
transformations or other such constructs from category theory. All we have here
|
|
|
|
|
is a regular maps on objects and a pair of arrows.
|
|
|
|
|
%
|
|
|
|
|
This data must satisfy:
|
|
|
|
|
%
|
|
|
|
|
\begin{align}
|
|
|
|
|
\label{eq:monad-monoidal-laws}
|
|
|
|
|
\begin{split}
|
|
|
|
|
\bind\ \pure & ≡ \identity_{\EndoR\ X}
|
|
|
|
|
\\
|
|
|
|
|
% \prod_{f \tp \Arrow\ X\ (\EndoR\ Y)}
|
|
|
|
|
\pure \fish f & ≡ f
|
|
|
|
|
\\
|
|
|
|
|
% \prod_{\substack{g \tp \Arrow\ Y\ (\EndoR\ Z)\\f \tp \Arrow\ X\ (\EndoR\ Y)}}
|
|
|
|
|
(\bind\ f) \rrr (\bind\ g) & ≡ \bind\ (f \fish g)
|
|
|
|
|
\end{split}
|
|
|
|
|
\end{align}
|
|
|
|
|
%
|
|
|
|
|
Here likewise the arrows $f \tp \Arrow\ X\ (\EndoR\ Y)$ and $g \tp
|
|
|
|
|
\Arrow\ Y\ (\EndoR\ Z)$ are universally quantified (as well as the objects they
|
|
|
|
|
range over). $\fish$ is the kleisli-arrow which is defined as $f \fish g \defeq
|
|
|
|
|
f \rrr (\bind\ g)$ . (\TODO{Better way to typeset $\fish$?})
|
|
|
|
|
|
|
|
|
|
%% Definition of what it means for a category to have all products.
|
|
|
|
|
\subsection{Equivalence of formulations}
|
|
|
|
|
%
|
|
|
|
|
In my implementation I proceede to show how the one formulation gives rise to
|
|
|
|
|
the other and vice-versa. For the present purpose I will briefly sketch some
|
|
|
|
|
parts of this construction:
|
|
|
|
|
|
|
|
|
|
%% \WIP{} Prove propositionality for being a product and having products.
|
|
|
|
|
The notation I have chosen here in the report
|
|
|
|
|
overloads e.g. $\pure$ to both refer to a natural transformation and an arrow.
|
|
|
|
|
This is of course not a coincidence as the arrow in the kleisli formulation
|
|
|
|
|
shall correspond exactly to the map on arrows from the natural transformation
|
|
|
|
|
called $\pure$.
|
|
|
|
|
|
|
|
|
|
%% \subsubsection{Exponentials}
|
|
|
|
|
%% Definition of what it means to be an exponential object.
|
|
|
|
|
In the monoidal formulation we can define $\bind$:
|
|
|
|
|
%
|
|
|
|
|
\begin{align}
|
|
|
|
|
\bind \defeq \join \lll \fmap\ f
|
|
|
|
|
\end{align}
|
|
|
|
|
%
|
|
|
|
|
And likewise in the kleisli formulation we can define $\join$:
|
|
|
|
|
%
|
|
|
|
|
\begin{align}
|
|
|
|
|
\join \defeq \bind\ \identity
|
|
|
|
|
\end{align}
|
|
|
|
|
%
|
|
|
|
|
Which shall play the role of $\join$.
|
|
|
|
|
|
|
|
|
|
%% Definition of what it means for a category to have all exponential objects.
|
|
|
|
|
|
|
|
|
|
%% \subsubsection{Cartesian closed categories}
|
|
|
|
|
%% Definition of what it means for a category to be cartesian closed; namely that
|
|
|
|
|
%% it has all products and all exponentials.
|
|
|
|
|
|
|
|
|
|
%% \subsubsection{Natural transformations}
|
|
|
|
|
%% Definition of transformations\footnote{Maybe this is a name I made up for a
|
|
|
|
|
%% family of morphisms} and the naturality condition for these.
|
|
|
|
|
|
|
|
|
|
%% Proof that naturality is a mere proposition and the accompanying equality
|
|
|
|
|
%% principle. Proof that natural transformations are homotopic sets.
|
|
|
|
|
|
|
|
|
|
%% The identity natural transformation.
|
|
|
|
|
|
|
|
|
|
%% \subsubsection{Yoneda embedding}
|
|
|
|
|
|
|
|
|
|
%% The yoneda embedding is typically presented in terms of the category of
|
|
|
|
|
%% categories (cf. Awodey) \emph however this is not stricly needed - all we need
|
|
|
|
|
%% is what would be the exponential object in that category - this happens to be
|
|
|
|
|
%% functors and so this is how we define the yoneda embedding.
|
|
|
|
|
|
|
|
|
|
%% \subsubsection{Monads}
|
|
|
|
|
|
|
|
|
|
%% Defines an equivalence between these two formulations of a monad:
|
|
|
|
|
|
|
|
|
|
%% \subsubsubsection{Monoidal monads}
|
|
|
|
|
|
|
|
|
|
%% Defines the standard monoidal representation of a monad:
|
|
|
|
|
|
|
|
|
|
%% An endofunctor with two natural transformations (called ``pure'' and ``join'')
|
|
|
|
|
%% and some laws about these natural transformations.
|
|
|
|
|
|
|
|
|
|
%% Propositionality proofs and equality principle is provided.
|
|
|
|
|
|
|
|
|
|
%% \subsubsubsection{Kleisli monads}
|
|
|
|
|
|
|
|
|
|
%% A presentation of monads perhaps more familiar to a functional programer:
|
|
|
|
|
|
|
|
|
|
%% A map on objects and two maps on morphisms (called ``pure'' and ``bind'') and
|
|
|
|
|
%% some laws about these maps.
|
|
|
|
|
|
|
|
|
|
%% Propositionality proofs and equality principle is provided.
|
|
|
|
|
|
|
|
|
|
%% \subsubsubsection{Voevodsky's construction}
|
|
|
|
|
|
|
|
|
|
%% Provides construction 2.3 as presented in an unpublished paper by Vladimir
|
|
|
|
|
%% Voevodsky. This construction is similiar to the equivalence provided for the two
|
|
|
|
|
%% preceding formulations
|
|
|
|
|
%% \footnote{ TODO: I would like to include in the thesis some motivation for why
|
|
|
|
|
%% this construction is particularly interesting.}
|
|
|
|
|
|
|
|
|
|
%% \subsubsection{Functors}
|
|
|
|
|
%% The category of functors and natural transformations. An immediate corrolary is
|
|
|
|
|
%% the set of presheaf categories.
|
|
|
|
|
|
|
|
|
|
%% \WIP{} I have not shown that the category of functors is univalent.
|
|
|
|
|
|
|
|
|
|
%% \subsubsection{Relations}
|
|
|
|
|
%% The category of relations. \WIP{} I have not shown that this category is
|
|
|
|
|
%% univalent. Not sure I intend to do so either.
|
|
|
|
|
|
|
|
|
|
%% \subsubsection{Free category}
|
|
|
|
|
%% The free category of a category. \WIP{} I have not shown that this category is
|
|
|
|
|
%% univalent.
|
|
|
|
|
|
|
|
|
|
%% \subsection{Current Challenges}
|
|
|
|
|
%% Besides the items marked \WIP{} above I still feel a bit unsure about what to
|
|
|
|
|
%% include in my report. Most of my work so far has been specifically about
|
|
|
|
|
%% developing this library. Some ideas:
|
|
|
|
|
%% %
|
|
|
|
|
%% \begin{itemize}
|
|
|
|
|
%% \item
|
|
|
|
|
%% Modularity properties
|
|
|
|
|
%% \item
|
|
|
|
|
%% Compare with setoid-approach to solve similiar problems.
|
|
|
|
|
%% \item
|
|
|
|
|
%% How to structure an implementation to best deal with types that have no
|
|
|
|
|
%% structure (propositions) and those that do (sets and everything above)
|
|
|
|
|
%% \end{itemize}
|
|
|
|
|
%% %
|
|
|
|
|
%% \subsection{Ideas for future developments}
|
|
|
|
|
%% \subsubsection{Higher categories}
|
|
|
|
|
%% I only have a notion of (1-)categories. Perhaps it would be nice to also
|
|
|
|
|
%% formalize higher categories.
|
|
|
|
|
|
|
|
|
|
%% \subsubsection{Hierarchy of concepts related to monads}
|
|
|
|
|
%% In Haskell the type-class Monad sits in a hierarchy atop the notion of a functor
|
|
|
|
|
%% and applicative functors. There's probably a similiar notion in the
|
|
|
|
|
%% category-theoretic approach to developing this.
|
|
|
|
|
|
|
|
|
|
%% As I have already defined monads from these two perspectives, it would be
|
|
|
|
|
%% interesting to take this idea even further and actually show how monads are
|
|
|
|
|
%% related to applicative functors and functors. I'm not entirely sure how this
|
|
|
|
|
%% would look in Agda though.
|
|
|
|
|
|
|
|
|
|
%% \subsubsection{Use formulation on the standard library}
|
|
|
|
|
%% I also thought it would be interesting to use this library to show certain
|
|
|
|
|
%% properties about functors, applicative functors and monads used in the Agda
|
|
|
|
|
%% Standard library. So I went ahead and tried to show that agda's standard
|
|
|
|
|
%% library's notion of a functor (along with suitable laws) is equivalent to my
|
|
|
|
|
%% formulation (in the category of homotopic sets). I ran into two problems here,
|
|
|
|
|
%% however; the first one is that the standard library's notion of a functor is
|
|
|
|
|
%% indexed by the object map:
|
|
|
|
|
%% %
|
|
|
|
|
%% $$
|
|
|
|
|
%% \Functor \tp (\Type \to \Type) \to \Type
|
|
|
|
|
%% $$
|
|
|
|
|
%% %
|
|
|
|
|
%% Where $\Functor\ F$ has the member:
|
|
|
|
|
%% %
|
|
|
|
|
%% $$
|
|
|
|
|
%% \fmap \tp (A \to B) \to F A \to F B
|
|
|
|
|
%% $$
|
|
|
|
|
%% %
|
|
|
|
|
%% Whereas the object map in my definition is existentially quantified:
|
|
|
|
|
%% %
|
|
|
|
|
%% $$
|
|
|
|
|
%% \Functor \tp \Type
|
|
|
|
|
%% $$
|
|
|
|
|
%% %
|
|
|
|
|
%% And $\Functor$ has these members:
|
|
|
|
|
%% \begin{align*}
|
|
|
|
|
%% F & \tp \Type \to \Type \\
|
|
|
|
|
%% \fmap & \tp (A \to B) \to F A \to F B\}
|
|
|
|
|
%% \end{align*}
|
|
|
|
|
%% %
|
|
|
|
|
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.
|
|
|
|
|
|
