Use unicode symbols
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@ -112,13 +112,13 @@ definitions: If we let $p$ be a witness to the identity law, which formally is:
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%
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\begin{equation}
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\label{eq:identity}
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\var{IsIdentity} \defeq
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\prod_{A, B \tp \Object} \prod_{f \tp \Arrow\ A\ B}
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\left(\id \lll f \equiv f\right) \x \left(f \lll \id \equiv f\right)
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\var{IsIdentity} ≜
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∏_{A, B \tp \Object} ∏_{f \tp \Arrow\ A\ B}
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\left(\id \lll f ≡ f\right) \x \left(f \lll \id ≡ f\right)
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\end{equation}
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%
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Then we can construct the identity isomorphism $\idIso \tp (\identity,
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\identity, p) \tp A \approxeq A$ for any object $A$. Here $\approxeq$
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\identity, p) \tp A ≊ A$ for any object $A$. Here $≊$
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denotes isomorphism on objects (whereas $\cong$ denotes isomorphism on
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types). This will be elaborated further on in sections
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\S\ref{sec:equiv} and \S\ref{sec:univalence}. Moreover due to
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@ -135,7 +135,7 @@ isomorphism on objects play the role of equivalence on types. Formally:
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%
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\begin{align}
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\label{eq:cat-univ}
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\isEquiv\ (A \equiv B)\ (A \approxeq B)\ \idToIso
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\isEquiv\ (A ≡ B)\ (A ≊ B)\ \idToIso
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\end{align}
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%
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Note that \ref{eq:cat-univ} is \emph{not} the same as:
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@ -143,7 +143,7 @@ Note that \ref{eq:cat-univ} is \emph{not} the same as:
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\begin{equation}
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\label{eq:cat-univalence}
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%% \tag{Univalence, category}
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(A \equiv B) \simeq (A \approxeq B)
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(A ≡ B) ≃ (A ≊ B)
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\end{equation}
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%
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However the two are logically equivalent: One can construct the latter
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@ -156,9 +156,9 @@ data:
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%
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\begin{align}
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\Object & \tp \Type \\
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\Arrow & \tp \Object \to \Object \to \Type \\
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\Arrow & \tp \Object → \Object → \Type \\
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\identity & \tp \Arrow\ A\ A \\
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\lll & \tp \Arrow\ B\ C \to \Arrow\ A\ B \to \Arrow\ A\ C
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\lll & \tp \Arrow\ B\ C → \Arrow\ A\ B → \Arrow\ A\ C
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\end{align}
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%
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And laws:
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@ -179,7 +179,7 @@ f \lll \identity ≡ f
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%% \tag{arrows are sets}
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\isSet\ (\Arrow\ A\ B)\\
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\tag{\ref{eq:cat-univ}}
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\isEquiv\ (A \equiv B)\ (A \approxeq B)\ \idToIso
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\isEquiv\ (A ≡ B)\ (A ≊ B)\ \idToIso
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\end{align}
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%
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$\lll$ denotes arrow composition (right-to-left), and reverse function
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@ -205,25 +205,25 @@ is by way of the `combinators' $\propPi$ and $\propSig$ presented in sections
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\S\ref{sec:propPi} and \S\ref{sec:propSig}:
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%
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\begin{align*}
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\propPi & \tp \left(\prod_{a \tp A} \isProp\ (P\ a)\right) \to \isProp\ \left(\prod_{a \tp A} P\ a\right)
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\propPi & \tp \left(∏_{a \tp A} \isProp\ (P\ a)\right) → \isProp\ \left(∏_{a \tp A} P\ a\right)
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\\
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\propSig & \tp \isProp\ A \to \left(\prod_{a \tp A} \isProp\ (P\ a)\right) \to \isProp\ \left(\sum_{a \tp A} P\ a\right)
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\propSig & \tp \isProp\ A → \left(∏_{a \tp A} \isProp\ (P\ a)\right) → \isProp\ \left(∑_{a \tp A} P\ a\right)
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\end{align*}
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%
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The proof goes like this: We `eliminate' the 3 function abstractions
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by applying $\propPi$ three times. So our proof obligation becomes:
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%
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$$
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\isProp\ \left( \left( \id \comp f \equiv f \right) \x \left( f \comp \id \equiv f \right) \right)
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\isProp\ \left( \left( \id \comp f ≡ f \right) \x \left( f \comp \id ≡ f \right) \right)
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$$
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%
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Then we eliminate the (non-dependent) sigma-type by applying $\propSig$ giving
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us the two obligations $\isProp\ (\id \comp f \equiv f)$ and $\isProp\ (f \comp
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\id \equiv f)$ which follows from the type of arrows being a
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us the two obligations $\isProp\ (\id \comp f ≡ f)$ and $\isProp\ (f \comp
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\id ≡ f)$ which follows from the type of arrows being a
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set.
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This example illustrates nicely how we can use these combinators to
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reason about `canonical' types like $\sum$ and $\prod$. Similar
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reason about `canonical' types like $∑$ and $∏$. Similar
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combinators can be defined at the other homotopic levels. These
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combinators are however not applicable in situations where we want to
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reason about other types e.g.\ types we have defined ourselves. For
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@ -236,7 +236,7 @@ show that the type of pre-categories is also a proposition. Formally:
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\isProp\ \IsPreCategory
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\end{equation}
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%
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Where The definition of $\IsPreCategory$ is the triple:
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Where the definition of $\IsPreCategory$ is the triple:
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%
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\begin{align*}
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\var{isAssociative} & \tp \var{IsAssociative}\\
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@ -252,12 +252,12 @@ the path type must be used directly.
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The type \ref{eq:propIsPreCategory} is judgmentally the same as:
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%
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$$
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\prod_{a, b \tp \IsPreCategory} a \equiv b
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∏_{a, b \tp \IsPreCategory} a ≡ b
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$$
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%
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So to prove the proposition let $a, b \tp \IsPreCategory$ be given. To
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prove the equality $a \equiv b$ is to give a continuous path from the
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index-type into the path-space. I.e.\ a function $\I \to
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prove the equality $a ≡ b$ is to give a continuous path from the
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index-type into the path-space. I.e.\ a function $\I →
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\IsPreCategory$. This path must satisfy being being judgmentally the
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same as $a$ at the left endpoint and $b$ at the right endpoint. We
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know we can form a continuous path between all projections of $a$ and
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@ -268,14 +268,14 @@ $b.\isIdentity$ is simply formed by:
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$$
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\propIsIdentity\ a.\isIdentity\ b.\isIdentity
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\tp
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a.\isIdentity \equiv b.\isIdentity
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a.\isIdentity ≡ b.\isIdentity
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$$
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%
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So to give the continuous function $\I \to \IsPreCategory$, which is our goal, we
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So to give the continuous function $\I → \IsPreCategory$, which is our goal, we
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introduce $i \tp \I$ and proceed by constructing an element of $\IsPreCategory$
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by using the fact that all the projections are propositions to generate paths
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between all projections. Once we have such a path e.g.\ $p \tp a.\isIdentity
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\equiv b.\isIdentity$ we can eliminate it with $i$ and thus obtain $p\ i \tp
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≡ b.\isIdentity$ we can eliminate it with $i$ and thus obtain $p\ i \tp
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(p\ i).\isIdentity$. This element satisfies exactly that it corresponds to the
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corresponding projections at either endpoint. Thus the element we construct at
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$i$ becomes the triple:
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@ -298,7 +298,7 @@ equalities at different levels. It is worth noting that proving this
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theorem with the regular inductive equality type would already not be
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possible, since we at least need functional
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extensionality\index{functional extensionality} (the projections are
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all $\prod$-types). Assuming we had functional extensionality
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all $∏$-types). Assuming we had functional extensionality
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available to us as an axiom, we would use functional extensionality
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\TODO{in reverse?} to retrieve the equalities in $a$ and $b$,
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pattern-match on them to see that they are both $\refl$ and then close
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@ -315,13 +315,13 @@ follow the same recipe as above, let $a, b \tp \IsCategory$ be given,
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to show them equal, we now need to give two paths. One homogeneous:
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%
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$$
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p \tp a.\isPreCategory \equiv b.\isPreCategory
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p \tp a.\isPreCategory ≡ b.\isPreCategory
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$$
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%
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and one heterogeneous:
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%
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$$
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\Path\ (\lambda\; i \to (p\ i).Univalent)\ a.\isPreCategory\ b.\isPreCategory
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\Path\ (\lambda\; i → (p\ i).Univalent)\ a.\isPreCategory\ b.\isPreCategory
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$$
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%
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Which depends on the choice of $p$. The first of these we can provide since, as
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@ -349,7 +349,7 @@ When we have a proper category we can make precise the notion of
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function:
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%
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$$
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\isoToId \tp (A \approxeq B) \to (A \equiv B)
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\isoToId \tp (A ≊ B) → (A ≡ B)
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$$
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%
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A perhaps somewhat surprising application of this is that we can show that
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@ -371,11 +371,11 @@ details. It is in the module:
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\section{Equivalences}
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\label{sec:equiv}
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The usual notion of a function $f \tp A \to B$ having an inverses is:
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The usual notion of a function $f \tp A → B$ having an inverses is:
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%
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\begin{equation}
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\label{eq:isomorphism}
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\sum_{g \tp B \to A} \left( f \comp g \equiv \identity \right) \x \left( g \comp f \equiv \identity \right)
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∑_{g \tp B → A} \left( f \comp g ≡ \identity \right) \x \left( g \comp f ≡ \identity \right)
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\end{equation}
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%
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This is defined in \cite[p. 129]{hott-2013} where it is referred to as the a
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@ -383,15 +383,15 @@ This is defined in \cite[p. 129]{hott-2013} where it is referred to as the a
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$\Isomorphism\ f$. This also gives rise to the following type:
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%
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\begin{equation}
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A \cong B \defeq \sum_{f \tp A \to B} \Isomorphism\ f
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A \cong B ≜ ∑_{f \tp A → B} \Isomorphism\ f
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\end{equation}
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%
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At the same place \cite{hott-2013} gives an ``interface'' for what the judgment
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$\isEquiv \tp (A \to B) \to \MCU$ must provide:
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$\isEquiv \tp (A → B) → \MCU$ must provide:
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%
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\begin{align}
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\var{fromIso} & \tp \Isomorphism\ f \to \isEquiv\ f \\
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\var{toIso} & \tp \isEquiv\ f \to \Isomorphism\ f \\
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\var{fromIso} & \tp \Isomorphism\ f → \isEquiv\ f \\
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\var{toIso} & \tp \isEquiv\ f → \Isomorphism\ f \\
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\label{eq:propIsEquiv}
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&\mathrel{\ } \isEquiv\ f
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\end{align}
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@ -399,8 +399,8 @@ $\isEquiv \tp (A \to B) \to \MCU$ must provide:
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The maps $\var{fromIso}$ and $\var{toIso}$ naturally extend to these maps:
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%
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\begin{align}
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\var{fromIsomorphism} & \tp A \cong B \to A \simeq B \\
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\var{toIsomorphism} & \tp A \simeq B \to A \cong B
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\var{fromIsomorphism} & \tp A \cong B → A ≃ B \\
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\var{toIsomorphism} & \tp A ≃ B → A \cong B
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\end{align}
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%
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Having this interface gives us both: a way to think rather abstractly about how
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@ -408,11 +408,11 @@ to work with equivalences and a way to use ad hoc definitions of equivalences.
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The specific instantiation of $\isEquiv$ as defined in \cite{cubical-agda} is:
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%
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$$
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isEquiv\ f \defeq \prod_{b \tp B} \isContr\ (\fiber\ f\ b)
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isEquiv\ f ≜ ∏_{b \tp B} \isContr\ (\fiber\ f\ b)
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$$
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where
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$$
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\fiber\ f\ b \defeq \sum_{a \tp A} \left( b \equiv f\ a \right)
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\fiber\ f\ b ≜ ∑_{a \tp A} \left( b ≡ f\ a \right)
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$$
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%
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I give its definition here mainly for completeness, because as I stated we can
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@ -430,16 +430,16 @@ equivalence between them:
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%
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\begin{equation}
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\label{eq:equivalence}
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A \simeq B \defeq \sum_{f \tp A \to B} \isEquiv\ f
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A ≃ B ≜ ∑_{f \tp A → B} \isEquiv\ f
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\end{equation}
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%
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Note that the term equivalence here is overloaded referring both to the map $f
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\tp A \to B$ and the type $A \simeq B$. The notion of an isomorphism is
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\tp A → B$ and the type $A ≃ B$. The notion of an isomorphism is
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similarly conflated as isomorphism can refer to the type $A \cong B$ as well as
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the the map $A \to B$ that witness this. I will use these conflated terms when
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the the map $A → B$ that witness this. I will use these conflated terms when
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it is clear from the context what is being referred to.
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Both $\cong$ and $\simeq$ form equivalence relations (no pun intended).
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Both $\cong$ and $≃$ form equivalence relations (no pun intended).
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\section{Univalence}
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\label{sec:univalence}
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@ -447,25 +447,25 @@ As noted in the introduction the univalence for types $A\; B \tp \Type$ states
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that:
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%
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$$
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\var{Univalence} \defeq (A \equiv B) \simeq (A \simeq B)
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\var{Univalence} ≜ (A ≡ B) ≃ (A ≃ B)
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$$
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%
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As mentioned the univalence criterion for some category $\bC$ says that for all
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\emph{objects} $A\;B$ we must have:
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$$
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\isEquiv\ (A \equiv B)\ (A \approxeq B)\ \idToIso
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\isEquiv\ (A ≡ B)\ (A ≊ B)\ \idToIso
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$$
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And I mentioned that this was logically equivalent to
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%
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$$
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(A \equiv B) \simeq (A \approxeq B)
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(A ≡ B) ≃ (A ≊ B)
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$$
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%
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Given that we saw in the previous section that we can construct an equivalence
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from an isomorphism it suffices to demonstrate:
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%
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$$
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(A \equiv B) \cong (A \approxeq B)
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(A ≡ B) \cong (A ≊ B)
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$$
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%
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That is, we must demonstrate that there is an isomorphism (on types) between
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@ -480,87 +480,87 @@ to behave similarly to maps.
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I will now mention a few helpful theorems that follow from univalence that will
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become useful later.
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Obviously univalence gives us an isomorphism between $A \equiv B$ and $A
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\approxeq B$. I will name these for convenience:
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Obviously univalence gives us an isomorphism between $A ≡ B$ and $A
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≊ B$. I will name these for convenience:
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%
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$$
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\idToIso \tp A \equiv B \to A \approxeq B
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\idToIso \tp A ≡ B → A ≊ B
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$$
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%
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$$
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\isoToId \tp A \approxeq B \to A \equiv B
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\isoToId \tp A ≊ B → A ≡ B
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$$
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%
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The next few theorems are variations on theorem 9.1.9 from \cite{hott-2013}. Let
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an isomorphism $A \approxeq B$ in some category $\bC$ be given. Name the
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isomorphism $\iota \tp A \to B$ and its inverse $\inv{\iota} \tp B \to A$.
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an isomorphism $A ≊ B$ in some category $\bC$ be given. Name the
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isomorphism $\iota \tp A → B$ and its inverse $\inv{\iota} \tp B → A$.
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Since $\bC$ is a category (and therefore univalent) the isomorphism induces a
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path $p \tp A \equiv B$. From this equality we can get two further paths:
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$p_{\var{dom}} \tp \Arrow\ A\ X \equiv \Arrow\ B\ X$ and
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$p_{\var{cod}} \tp \Arrow\ X\ A \equiv \Arrow\ X\ B$. We
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path $p \tp A ≡ B$. From this equality we can get two further paths:
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$p_{\var{dom}} \tp \Arrow\ A\ X ≡ \Arrow\ B\ X$ and
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$p_{\var{cod}} \tp \Arrow\ X\ A ≡ \Arrow\ X\ B$. We
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then have the following two theorems:
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%
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\begin{align}
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\label{eq:coeDom}
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\var{coeDom} & \tp \prod_{f \tp A \to X}
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\var{coe}\ p_{\var{dom}}\ f \equiv f \lll \inv{\iota}
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\var{coeDom} & \tp ∏_{f \tp A → X}
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\var{coe}\ p_{\var{dom}}\ f ≡ f \lll \inv{\iota}
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\\
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\label{eq:coeCod}
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\var{coeCod} & \tp \prod_{f \tp A \to X}
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\var{coe}\ p_{\var{cod}}\ f \equiv \iota \lll f
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\var{coeCod} & \tp ∏_{f \tp A → X}
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\var{coe}\ p_{\var{cod}}\ f ≡ \iota \lll f
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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 analogous.
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%
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\begin{align*}
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\var{coe}\ p_{\var{dom}}\ f
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& \equiv f \lll \inv{(\idToIso\ p)} && \text{lemma} \\
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& \equiv f \lll \inv{\iota}
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& ≡ f \lll \inv{(\idToIso\ p)} && \text{lemma} \\
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& ≡ f \lll \inv{\iota}
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&& \text{$\idToIso$ and $\isoToId$ are inverses}\\
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\end{align*}
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%
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In the second step we use the fact that $p$ is constructed from the isomorphism
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$\iota$ -- $\inv{(\idToIso\ p)}$ denotes the map $B \to A$ induced by the
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$\iota$ -- $\inv{(\idToIso\ p)}$ denotes the map $B → A$ induced by the
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isomorphism $\idToIso\ p \tp A \cong B$. The helper-lemma is similar to
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what we are trying to prove but talks about paths rather than isomorphisms:
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%
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\begin{equation}
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\label{eq:coeDomIso}
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\prod_{f \tp \Arrow\ A\ B} \prod_{p \tp A \equiv B}
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\var{coe}\ p_{\var{dom}}\ f \equiv f \lll \inv{(\idToIso\ p)}
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∏_{f \tp \Arrow\ A\ B} ∏_{p \tp A ≡ B}
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\var{coe}\ p_{\var{dom}}\ f ≡ f \lll \inv{(\idToIso\ p)}
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\end{equation}
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%
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Again $p_{\var{dom}}$ denotes the path $\Arrow\ A\ X \equiv
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Again $p_{\var{dom}}$ denotes the path $\Arrow\ A\ X ≡
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\Arrow\ B\ X$ induced by $p$. To prove this statement I let $f$ and $p$ be
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given and then invoke based-path-induction. The induction will be based at $A
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\tp \Object$ Let $\widetilde{B} \tp \Object$ and $\widetilde{p} \tp A
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\equiv \widetilde{B}$ be given. The family that we perform induction over will
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≡ \widetilde{B}$ be given. The family that we perform induction over will
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be:
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%
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\begin{align}
|
||||
D\ \widetilde{B}\ \widetilde{p} \defeq
|
||||
%% \prod_{\widetilde{B} \tp \Object}
|
||||
%% \prod_{\widetilde{p} \tp A \equiv \widetilde{B}}
|
||||
D\ \widetilde{B}\ \widetilde{p} ≜
|
||||
%% ∏_{\widetilde{B} \tp \Object}
|
||||
%% ∏_{\widetilde{p} \tp A ≡ \widetilde{B}}
|
||||
\var{coe}\ {\widetilde{p}}^*\ f
|
||||
\equiv
|
||||
≡
|
||||
f \lll \inv{(\idToIso\ \widetilde{p})}
|
||||
\end{align}
|
||||
The base-case therefore becomes
|
||||
$d \tp \var{coe}\ \refl^*\ f \equiv f \lll \inv{(\idToIso\ \refl)}$
|
||||
$d \tp \var{coe}\ \refl^*\ f ≡ f \lll \inv{(\idToIso\ \refl)}$
|
||||
and is inhabited by:
|
||||
\begin{align*}
|
||||
\var{coe}\ \refl^*\ f
|
||||
& \equiv f
|
||||
& ≡ f
|
||||
&& \text{$\refl$ is a neutral element for $\var{coe}$}\\
|
||||
& \equiv f \lll \identity \\
|
||||
& \equiv f \lll \var{subst}\ \refl\ \identity
|
||||
& ≡ f \lll \identity \\
|
||||
& ≡ f \lll \var{subst}\ \refl\ \identity
|
||||
&& \text{$\refl$ is a neutral element for $\var{subst}$}\\
|
||||
& \equiv f \lll \inv{(\idToIso\ \refl)}
|
||||
& ≡ f \lll \inv{(\idToIso\ \refl)}
|
||||
&& \text{By definition of $\idToIso$}\\
|
||||
\end{align*}
|
||||
%
|
||||
To close the based-path-induction we must supply the value ``at the other''. In
|
||||
this case this is simply $B \tp \Object$ and $p \tp A \equiv B$ which we have.
|
||||
this case this is simply $B \tp \Object$ and $p \tp A ≡ B$ which we have.
|
||||
In summary the proof of \ref{eq:coeDomIso} is the term:
|
||||
%
|
||||
\begin{equation}
|
||||
|
|
@ -594,14 +594,14 @@ category.
|
|||
Showing that this forms a pre-category is rather straightforward.
|
||||
%
|
||||
$$
|
||||
h \rrr (g \rrr f) \equiv h \rrr g \rrr f
|
||||
h \rrr (g \rrr f) ≡ h \rrr g \rrr f
|
||||
$$
|
||||
%
|
||||
Since $\rrr$ is reverse function composition this is just the symmetric version
|
||||
of associativity.
|
||||
%
|
||||
$$
|
||||
\identity \rrr f \equiv f \x f \rrr identity \equiv f
|
||||
\identity \rrr f ≡ f \x f \rrr \identity ≡ f
|
||||
$$
|
||||
%
|
||||
This is just the swapped version of identity.
|
||||
|
|
@ -613,21 +613,21 @@ arguments. Or in other words; since $\Arrow\ A\ B$ is a set for all $A\;B \tp
|
|||
Now, to show that this category is univalent is not as straightforward. Luckily
|
||||
section \S\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
|
||||
$\wideoverbar{\idToIso} \tp (A \equiv B) \to (A \wideoverbar{\approxeq} B)$.
|
||||
From the original category we have that $\idToIso \tp (A \equiv B) \to (A \cong
|
||||
$\wideoverbar{\idToIso} \tp (A ≡ B) → (A \wideoverbar{≊} B)$.
|
||||
From the original category we have that $\idToIso \tp (A ≡ B) → (A \cong
|
||||
B)$ is an isomorphism. Let us denote its inverse with $\isoToId \tp (A
|
||||
\approxeq B) \to (A \equiv B)$. If we squint we can see what we need is a way to
|
||||
go between $\wideoverbar{\approxeq}$ and $\approxeq$.
|
||||
≊ B) → (A ≡ B)$. If we squint we can see what we need is a way to
|
||||
go between $\wideoverbar{≊}$ and $≊$.
|
||||
|
||||
An inhabitant of $A \approxeq B$ is simply an arrow $f \tp \Arrow\ A\ B$
|
||||
An inhabitant of $A ≊ B$ is simply an arrow $f \tp \Arrow\ A\ B$
|
||||
and its inverse $g \tp \Arrow\ B\ A$. In the opposite category $g$ will
|
||||
play the role of the isomorphism and $f$ will be the inverse. Similarly we can
|
||||
go in the opposite direction. I name these maps $\shufflef \tp (A \approxeq
|
||||
B) \to (A \wideoverbar{\approxeq} B)$ and $\shufflef^{-1} \tp (A
|
||||
\wideoverbar{\approxeq} B) \to (A \approxeq B)$ respectively.
|
||||
go in the opposite direction. I name these maps $\shufflef \tp (A ≊
|
||||
B) → (A \wideoverbar{≊} B)$ and $\shufflef^{-1} \tp (A
|
||||
\wideoverbar{≊} B) → (A ≊ B)$ respectively.
|
||||
|
||||
As the inverse of $\wideoverbar{\idToIso}$ I will pick $\wideoverbar{\isoToId}
|
||||
\defeq \isoToId \comp \shufflef$. The proof that they are inverses go as
|
||||
≜ \isoToId \comp \shufflef$. The proof that they are inverses go as
|
||||
follows:
|
||||
%
|
||||
\begin{align*}
|
||||
|
|
@ -636,21 +636,21 @@ follows:
|
|||
\\
|
||||
%% ≡⟨ cong (λ \; φ → φ x) (cong (λ \; φ → η ⊙ shuffle ⊙ φ) (funExt lem)) ⟩ \\
|
||||
%
|
||||
& \equiv
|
||||
& ≡
|
||||
\isoToId \comp \shufflef \comp \inv{\shufflef} \comp \idToIso
|
||||
&& \text{lemma} \\
|
||||
%% ≡⟨⟩ \\
|
||||
& \equiv
|
||||
& ≡
|
||||
\isoToId \comp \idToIso
|
||||
&& \text{$\shufflef$ is an isomorphism} \\
|
||||
& \equiv
|
||||
& ≡
|
||||
\identity
|
||||
&& \text{$\isoToId$ is an isomorphism}
|
||||
\end{align*}
|
||||
%
|
||||
The other direction is analogous.
|
||||
|
||||
The lemma used in step 2 of this proof states that $\wideoverbar{idToIso} \equiv
|
||||
The lemma used in step 2 of this proof states that $\wideoverbar{idToIso} ≡
|
||||
\inv{\shufflef} \comp \idToIso$. This is a rather straightforward proof
|
||||
since being-an-inverse-of is a proposition, so it suffices to show that their
|
||||
first components are equal, but this holds judgmentally.
|
||||
|
|
@ -659,7 +659,7 @@ 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:
|
||||
%
|
||||
$$
|
||||
\prod_{\bC \tp \Category} \left(\bC^{\var{Op}}\right)^{\var{Op}} \equiv \bC
|
||||
∏_{\bC \tp \Category} \left(\bC^{\var{Op}}\right)^{\var{Op}} ≡ \bC
|
||||
$$
|
||||
%
|
||||
As we have seen the laws in $\left(\bC^{\var{Op}}\right)^{\var{Op}}$ get quite
|
||||
|
|
@ -671,7 +671,7 @@ just by giving an equality on the data-part. So, given a category $\bC$ all we
|
|||
must provide is the following proof:
|
||||
%
|
||||
$$
|
||||
\var{raw}\ \left(\bC^{\var{Op}}\right)^{\var{Op}} \equiv \var{raw}\ \bC
|
||||
\var{raw}\ \left(\bC^{\var{Op}}\right)^{\var{Op}} ≡ \var{raw}\ \bC
|
||||
$$
|
||||
%
|
||||
And these are judgmentally the same. I remind the reader that the left-hand side
|
||||
|
|
@ -681,7 +681,7 @@ 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:
|
||||
%
|
||||
$$\Set \defeq \sum_{A \tp \MCU} \isSet\ A$$
|
||||
$$\Set ≜ ∑_{A \tp \MCU} \isSet\ A$$
|
||||
%
|
||||
The more straightforward notion of a category where the objects are types is
|
||||
not a valid \mbox{(1-)category}. This stems from the fact that types in cubical
|
||||
|
|
@ -689,15 +689,15 @@ Agda types can have higher homotopic structure.
|
|||
|
||||
Univalence does not follow immediately from univalence for types:
|
||||
%
|
||||
$$(A \equiv B) \simeq (A \simeq B)$$
|
||||
$$(A ≡ B) ≃ (A ≃ B)$$
|
||||
%
|
||||
Because here $A, B \tp \Type$ whereas the objects in this category have the type
|
||||
$\Set$ so we cannot form the type $\var{hA} \simeq \var{hB}$ for objects
|
||||
$\Set$ so we cannot form the type $\var{hA} ≃ \var{hB}$ for objects
|
||||
$\var{hA}\;\var{hB} \tp \Set$. In stead I show that this category
|
||||
satisfies:
|
||||
%
|
||||
$$
|
||||
(\var{hA} \equiv \var{hB}) \simeq (\var{hA} \approxeq \var{hB})
|
||||
(\var{hA} ≡ \var{hB}) ≃ (\var{hA} ≊ \var{hB})
|
||||
$$
|
||||
%
|
||||
Which, as we saw in section \S\ref{sec:univalence}, is sufficient to show that the
|
||||
|
|
@ -706,18 +706,18 @@ 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) && \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}}
|
||||
((A, s_A) ≡ (B, s_B))
|
||||
& ≃ (A ≡ B) && \ref{eq:equivPropSig} \\
|
||||
& ≃ (A ≃ B) && \text{Univalence} \\
|
||||
& ≃ ((A, s_A) ≊ (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
|
||||
And since $≃$ 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)
|
||||
\left(∏_{a \tp A} \isProp\ (P\ a)\right) → ∏_{x\;y \tp ∑_{a \tp A} P\ a} (x ≡ y) ≃ (\fst\ x ≡ \fst\ y)
|
||||
\end{align}
|
||||
%
|
||||
The lemma states that for pairs whose second component are mere propositions
|
||||
|
|
@ -728,30 +728,30 @@ 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
|
||||
∏_{a \tp A} \left( P\ a ≃ Q\ a \right) → ∑_{a \tp A} P\ a ≃ ∑_{a \tp A} Q\ a
|
||||
\end{align}
|
||||
%
|
||||
Which says that if two type-families are equivalent at all points, then pairs
|
||||
with identical first components and these families as second components will
|
||||
also be equivalent. For our purposes $P \defeq \isEquiv\ A\ B$ and $Q \defeq
|
||||
also be equivalent. For our purposes $P ≜ \isEquiv\ A\ B$ and $Q ≜
|
||||
\Isomorphism$. So we must finally prove:
|
||||
%
|
||||
\begin{align}
|
||||
\label{eq:equivIso}
|
||||
\prod_{f \tp A \to B} \left( \isEquiv\ A\ B\ f \simeq \Isomorphism\ f \right)
|
||||
∏_{f \tp A → B} \left( \isEquiv\ A\ B\ f ≃ \Isomorphism\ f \right)
|
||||
\end{align}
|
||||
|
||||
First, lets prove \ref{eq:equivPropSig}: Let $propP \tp \prod_{a \tp A} \isProp\ (P\ a)$ and $x\;y \tp \sum_{a \tp A} P\ a$ be given. Because
|
||||
First, lets prove \ref{eq:equivPropSig}: Let $propP \tp ∏_{a \tp A} \isProp\ (P\ a)$ and $x\;y \tp ∑_{a \tp A} P\ a$ be given. Because
|
||||
of $\var{fromIsomorphism}$ it suffices to give an isomorphism between
|
||||
$x \equiv y$ and $\fst\ x \equiv \fst\ y$:
|
||||
$x ≡ y$ and $\fst\ x ≡ \fst\ y$:
|
||||
%
|
||||
%% FIXME: Too much alignement?
|
||||
\begin{equation*}
|
||||
\begin{aligned}
|
||||
f & \defeq \congruence\ \fst
|
||||
&& \tp x \equiv y && \to \fst\ x \equiv \fst\ y \\
|
||||
g & \defeq \var{lemSig}\ \var{propP}\ x\ y
|
||||
&& \tp \fst\ x \equiv \fst\ y && \to x \equiv y
|
||||
f & ≜ \congruence\ \fst
|
||||
&& \tp x ≡ y && → \fst\ x ≡ \fst\ y \\
|
||||
g & ≜ \var{lemSig}\ \var{propP}\ x\ y
|
||||
&& \tp \fst\ x ≡ \fst\ y && → x ≡ y
|
||||
\end{aligned}
|
||||
\end{equation*}
|
||||
%
|
||||
|
|
@ -761,9 +761,9 @@ suffices to give a path between its first components to construct an equality of
|
|||
the two pairs:
|
||||
%
|
||||
\begin{align*}
|
||||
\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
|
||||
\var{lemSig} \tp \left( ∏_{x \tp A} \isProp\ (B\ x) \right) →
|
||||
∏_{u\; v \tp ∑_{a \tp A} B\ a}
|
||||
\left( \fst\ u ≡ \fst\ v \right) → u ≡ v
|
||||
\end{align*}
|
||||
%
|
||||
The proof that these are indeed inverses has been omitted. The details
|
||||
|
|
@ -772,41 +772,41 @@ can be found in the module:
|
|||
\sourcelink{Cat.Categories.Sets}
|
||||
\end{center}
|
||||
|
||||
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
|
||||
Now to prove \ref{eq:equivSig}: Let $e \tp ∏_{a \tp A} \left( P\ a
|
||||
≃ Q\ a \right)$ be given. To prove the equivalence, it suffices
|
||||
to give an isomorphism between $∑_{a \tp A} P\ a$ and $∑_{a \tp
|
||||
A} Q\ a$, 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
|
||||
Lastly we prove \ref{eq:equivIso}. Let $f \tp A → B$ be given. For the maps we
|
||||
choose:
|
||||
%
|
||||
\begin{align*}
|
||||
\var{toIso}
|
||||
& \tp \isEquiv\ f \to \Isomorphism\ f \\
|
||||
& \tp \isEquiv\ f → \Isomorphism\ f \\
|
||||
\var{fromIso}
|
||||
& \tp \Isomorphism\ f \to \isEquiv\ f
|
||||
& \tp \Isomorphism\ f → \isEquiv\ f
|
||||
\end{align*}
|
||||
%
|
||||
As mentioned in section \S\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*}
|
||||
\var{fromIso} \comp \var{toIso} \equiv \identity_{\isEquiv\ f}
|
||||
\var{fromIso} \comp \var{toIso} ≡ \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*}
|
||||
\var{toIso} \comp \var{fromIso} \equiv \identity_{\Isomorphism\ f}
|
||||
\var{toIso} \comp \var{fromIso} ≡ \identity_{\Isomorphism\ f}
|
||||
\end{align*}
|
||||
%
|
||||
We will show that $\Isomorphism\ f$ is also a mere proposition. To this
|
||||
end, let $X\;Y \tp \Isomorphism\ f$ 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 $\Path\ (\lambda\; i \to
|
||||
→ A$ respectively. Now, the proof that $X$ and $Y$ are the same is a pair of
|
||||
paths: $p \tp x ≡ y$ and $\Path\ (\lambda\; i →
|
||||
\var{AreInverses}\ f\ (p\ i))\ \mathcal{X}\ \mathcal{Y}$ where $\mathcal{X}$
|
||||
and $\mathcal{Y}$ denotes the witnesses that $x$ (respectively $y$) is an
|
||||
inverse to $f$. The path $p$ is inhabited by:
|
||||
|
|
@ -814,10 +814,10 @@ inverse to $f$. The path $p$ is inhabited by:
|
|||
\begin{align*}
|
||||
x
|
||||
& = x \comp \identity \\
|
||||
& \equiv x \comp (f \comp y)
|
||||
& ≡ x \comp (f \comp y)
|
||||
&& \text{$y$ is an inverse to $f$} \\
|
||||
& \equiv (x \comp f) \comp y \\
|
||||
& \equiv \identity \comp y
|
||||
& ≡ (x \comp f) \comp y \\
|
||||
& ≡ \identity \comp y
|
||||
&& \text{$x$ is an inverse to $f$} \\
|
||||
& = y
|
||||
\end{align*}
|
||||
|
|
@ -827,7 +827,7 @@ proposition and then use $\lemPropF$. So we prove the generalization:
|
|||
%
|
||||
\begin{align}
|
||||
\label{eq:propAreInversesGen}
|
||||
\prod_{g \tp B \to A} \isProp\ (\var{AreInverses}\ f\ g)
|
||||
∏_{g \tp B → A} \isProp\ (\var{AreInverses}\ f\ g)
|
||||
\end{align}
|
||||
%
|
||||
But $\var{AreInverses}\ f\ g$ is a pair of equations on arrows, so we use
|
||||
|
|
@ -854,7 +854,7 @@ demonstrated. The goal in this section is to show that products are
|
|||
propositions:
|
||||
%
|
||||
$$
|
||||
\prod_{\bC \tp \Category} \prod_{A\;B \tp \Object} \isProp\ (\var{Product}\ \bC\ A\ B)
|
||||
∏_{\bC \tp \Category} ∏_{A\;B \tp \Object} \isProp\ (\var{Product}\ \bC\ A\ B)
|
||||
$$
|
||||
%
|
||||
Where $\var{Product}\ \bC\ A\ B$ denotes the type of products of objects $A$
|
||||
|
|
@ -867,7 +867,7 @@ we recall the definition of products.
|
|||
\subsection{Definition of products}
|
||||
Given a category $\bC$ and two objects $A$ and $B$ in $\bC$ we define the
|
||||
product (object) 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
|
||||
$\pi_1 \tp A \x B → A$ and $\pi_2 \tp A \x B → B$ 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
|
||||
|
|
@ -875,9 +875,9 @@ 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}\\
|
||||
%% ∏_{X \tp Object} ∏_{f \tp \Arrow\ X\ A} ∏_{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
|
||||
\pi_1 \lll \pi ≡ f \x \pi_2 \lll \pi ≡ g
|
||||
\end{align}
|
||||
%
|
||||
$\pi$ is called the product (arrow) of $f$ and $g$.
|
||||
|
|
@ -900,8 +900,8 @@ category will consist of an arrow from the underlying category $\pairf \tp
|
|||
%
|
||||
\begin{align}
|
||||
\label{eq:pairArrowLaw}
|
||||
b_0 \lll f \equiv a_0 \x
|
||||
b_1 \lll f \equiv a_1
|
||||
b_0 \lll f ≡ a_0 \x
|
||||
b_1 \lll f ≡ a_1
|
||||
\end{align}
|
||||
|
||||
The identity morphism is the identity morphism from the underlying category.
|
||||
|
|
@ -914,13 +914,13 @@ choose $g \lll f$ and we must now verify that it satisfies
|
|||
%
|
||||
\begin{align*}
|
||||
c_0 \lll (f \lll g)
|
||||
& \equiv
|
||||
& ≡
|
||||
(c_0 \lll f) \lll g
|
||||
&& \text{Associativity} \\
|
||||
& \equiv
|
||||
& ≡
|
||||
b_0 \lll g
|
||||
&& \text{$f$ satisfies \ref{eq:pairArrowLaw}} \\
|
||||
& \equiv
|
||||
& ≡
|
||||
a_0
|
||||
&& \text{$g$ satisfies \ref{eq:pairArrowLaw}} \\
|
||||
\end{align*}
|
||||
|
|
@ -932,7 +932,7 @@ arrows form a set. For instance, to prove associativity we must prove that
|
|||
\begin{align}
|
||||
\label{eq:productAssoc}
|
||||
\overline{h} \lll (\overline{g} \lll \overline{f})
|
||||
\equiv
|
||||
≡
|
||||
(\overline{h} \lll \overline{g}) \lll \overline{f}
|
||||
\end{align}
|
||||
%
|
||||
|
|
@ -947,7 +947,7 @@ The proof obligations is consists of two things. The first one is:
|
|||
\begin{align}
|
||||
\label{eq:productAssocUnderlying}
|
||||
h \lll (g \lll f)
|
||||
\equiv
|
||||
≡
|
||||
(h \lll g) \lll f
|
||||
\end{align}
|
||||
%
|
||||
|
|
@ -976,7 +976,7 @@ stead we generalize \ref{eq:productPath} to:
|
|||
%
|
||||
\begin{align}
|
||||
\label{eq:productEqPrinc}
|
||||
\prod_{f \tp \Arrow\ X\ Y} \isProp\ \left( y_{\pairA} \lll f ≡ x_{\pairA} × y_{\pairB} \lll f ≡ x_{\pairB} \right)
|
||||
∏_{f \tp \Arrow\ X\ Y} \isProp\ \left( y_{\pairA} \lll f ≡ x_{\pairA} × y_{\pairB} \lll f ≡ x_{\pairB} \right)
|
||||
\end{align}
|
||||
%
|
||||
For all objects $X , x_{\pairA} , x_{\pairB}$ and $Y , y_{\pairA} , y_{\pairB}$,
|
||||
|
|
@ -986,7 +986,7 @@ in this category that says that to prove two arrows $f, f_0, f_1$ and $g, g_0,
|
|||
g_1$ equal it suffices to give a proof that $f$ and $g$ are equal.
|
||||
%% %
|
||||
%% $$
|
||||
%% \prod_{(f, f_0, f_1)\; (g,g_0,g_1) \tp \Arrow\ X\ Y} f \equiv g \to (f, f_0, f_1) \equiv (g,g_0,g_1)
|
||||
%% ∏_{(f, f_0, f_1)\; (g,g_0,g_1) \tp \Arrow\ X\ Y} f ≡ g \to (f, f_0, f_1) ≡ (g,g_0,g_1)
|
||||
%% $$
|
||||
%% %
|
||||
And thus we have proven \ref{eq:productAssoc} simply with
|
||||
|
|
@ -1007,7 +1007,7 @@ $$
|
|||
Which holds. And
|
||||
%
|
||||
$$
|
||||
\prod_{f \tp \Arrow\ X\ Y}
|
||||
∏_{f \tp \Arrow\ X\ Y}
|
||||
\isSet\ \left( y_{\pairA} \lll f ≡ x_{\pairA}
|
||||
× y_{\pairB} \lll f ≡ x_{\pairB}
|
||||
\right)
|
||||
|
|
@ -1024,7 +1024,7 @@ That is, for any two objects $\mathcal{X} = (X, x_{\mathcal{A}} , x_{\mathcal{B}
|
|||
and $\mathcal{Y} = Y, y_{\mathcal{A}}, y_{\mathcal{B}}$ I will show:
|
||||
%
|
||||
\begin{align}
|
||||
(\mathcal{X} \equiv \mathcal{Y}) \cong (\mathcal{X} \approxeq \mathcal{Y})
|
||||
(\mathcal{X} ≡ \mathcal{Y}) \cong (\mathcal{X} ≊ \mathcal{Y})
|
||||
\end{align}
|
||||
|
||||
I do this by showing that the following sequence of types are isomorphic.
|
||||
|
|
@ -1041,7 +1041,7 @@ The next types will be the triple:
|
|||
\begin{align}
|
||||
\label{eq:univ-1}
|
||||
\begin{split}
|
||||
p \tp & X \equiv Y \\
|
||||
p \tp & X ≡ Y \\
|
||||
& \Path\ (λ \; i → \Arrow\ (p\ i)\ \mathcal{A})\ x_{\mathcal{A}}\ y_{\mathcal{A}} \\
|
||||
& \Path\ (λ \; i → \Arrow\ (p\ i)\ \mathcal{B})\ x_{\mathcal{B}}\ y_{\mathcal{B}}
|
||||
\end{split}
|
||||
|
|
@ -1054,13 +1054,13 @@ isomorphism, and create a path from this:
|
|||
\begin{align}
|
||||
\label{eq:univ-2}
|
||||
\begin{split}
|
||||
\var{iso} \tp & X \approxeq Y \\
|
||||
\var{iso} \tp & X ≊ Y \\
|
||||
& \Path\ (λ \; i → \Arrow\ (\widetilde{p}\ i)\ \mathcal{A})\ x_{\mathcal{A}}\ y_{\mathcal{A}} \\
|
||||
& \Path\ (λ \; i → \Arrow\ (\widetilde{p}\ i)\ \mathcal{B})\ x_{\mathcal{B}}\ y_{\mathcal{B}}
|
||||
\end{split}
|
||||
\end{align}
|
||||
%
|
||||
Where $\widetilde{p} \defeq \isoToId\ \var{iso} \tp X \equiv Y$.
|
||||
Where $\widetilde{p} ≜ \isoToId\ \var{iso} \tp X ≡ Y$.
|
||||
|
||||
Finally we have the type:
|
||||
%
|
||||
|
|
@ -1085,15 +1085,15 @@ This proof of this has been omitted but can be found in the module:
|
|||
will show two corollaries of \ref{eq:coeCod}: For an isomorphism $(\iota,
|
||||
\inv{\iota}, \var{inv}) \tp A \cong B$, arrows $f \tp \Arrow\ A\ X$, $g \tp
|
||||
\Arrow\ B\ X$ and a heterogeneous path between them, $q \tp \Path\ (\lambda\; i
|
||||
\to p_{\var{dom}}\ i)\ f\ g$, where $p_{\var{dom}} \tp \Arrow\ A\ X \equiv
|
||||
→ p_{\var{dom}}\ i)\ f\ g$, where $p_{\var{dom}} \tp \Arrow\ A\ X ≡
|
||||
\Arrow\ B\ X$ is a path induced by $\var{iso}$, we have the following two
|
||||
results
|
||||
%
|
||||
\begin{align}
|
||||
\label{eq:domain-twist-0}
|
||||
f & \equiv g \lll \iota \\
|
||||
f & ≡ g \lll \iota \\
|
||||
\label{eq:domain-twist-1}
|
||||
g & \equiv f \lll \inv{\iota}
|
||||
g & ≡ f \lll \inv{\iota}
|
||||
\end{align}
|
||||
%
|
||||
The proof is omitted but can be found in the module:
|
||||
|
|
@ -1105,14 +1105,14 @@ Now we can prove the equivalence in the following way: Given $(f, \inv{f},
|
|||
\var{inv}_f) \tp X \cong Y$ and two heterogeneous paths
|
||||
%
|
||||
\begin{align*}
|
||||
p_{\mathcal{A}} & \tp \Path\ (\lambda\; i \to p_{\var{dom}}\ i)\ x_{\mathcal{A}}\ y_{\mathcal{A}}\\
|
||||
p_{\mathcal{A}} & \tp \Path\ (\lambda\; i → p_{\var{dom}}\ i)\ x_{\mathcal{A}}\ y_{\mathcal{A}}\\
|
||||
%
|
||||
q_{\mathcal{B}} & \tp \Path\ (\lambda\; i \to p_{\var{dom}}\ i)\ x_{\mathcal{B}}\ y_{\mathcal{B}}
|
||||
q_{\mathcal{B}} & \tp \Path\ (\lambda\; i → p_{\var{dom}}\ i)\ x_{\mathcal{B}}\ y_{\mathcal{B}}
|
||||
\end{align*}
|
||||
%
|
||||
all as in \ref{eq:univ-2}. I use $p_{\var{dom}}$ here again to mean the path
|
||||
induced by the isomorphism $f, \inv{f}$. I must now construct an isomorphism
|
||||
$(X, x_{\mathcal{A}}, x_{\mathcal{B}}) \approxeq (Y, y_{\mathcal{A}}, y_{\mathcal{B}})$
|
||||
$(X, x_{\mathcal{A}}, x_{\mathcal{B}}) ≊ (Y, y_{\mathcal{A}}, y_{\mathcal{B}})$
|
||||
as in \ref{eq:univ-3}. That is, an isomorphism in the present category. I remind
|
||||
the reader that such a gadget is a triple. The first component shall be:
|
||||
%
|
||||
|
|
@ -1145,7 +1145,7 @@ For the other direction we are given the isomorphism:
|
|||
$$
|
||||
(f, \inv{f}, \var{inv}_f, \var{inv}_{\inv{f}})
|
||||
\tp
|
||||
(X, x_{\mathcal{A}}, x_{\mathcal{B}}) \approxeq (Y, y_{\mathcal{A}}, y_{\mathcal{B}})
|
||||
(X, x_{\mathcal{A}}, x_{\mathcal{B}}) ≊ (Y, y_{\mathcal{A}}, y_{\mathcal{B}})
|
||||
$$
|
||||
%
|
||||
Projecting out the first component gives us the isomorphism
|
||||
|
|
@ -1155,16 +1155,16 @@ $$
|
|||
, \congruence\ \fst\ \var{inv}_f
|
||||
, \congruence\ \fst\ \var{inv}_{\inv{f}}
|
||||
)
|
||||
\tp X \approxeq Y
|
||||
\tp X ≊ Y
|
||||
$$
|
||||
%
|
||||
This gives rise to the following paths:
|
||||
%
|
||||
\begin{align}
|
||||
\begin{split}
|
||||
\widetilde{p} & \tp X \equiv Y \\
|
||||
\widetilde{p}_{\mathcal{A}} & \tp \Arrow\ X\ \mathcal{A} \equiv \Arrow\ Y\ \mathcal{A} \\
|
||||
\widetilde{p}_{\mathcal{B}} & \tp \Arrow\ X\ \mathcal{B} \equiv \Arrow\ Y\ \mathcal{B}
|
||||
\widetilde{p} & \tp X ≡ Y \\
|
||||
\widetilde{p}_{\mathcal{A}} & \tp \Arrow\ X\ \mathcal{A} ≡ \Arrow\ Y\ \mathcal{A} \\
|
||||
\widetilde{p}_{\mathcal{B}} & \tp \Arrow\ X\ \mathcal{B} ≡ \Arrow\ Y\ \mathcal{B}
|
||||
\end{split}
|
||||
\end{align}
|
||||
%
|
||||
|
|
@ -1181,7 +1181,7 @@ It then remains to construct the two paths:
|
|||
This is achieved with the following lemma:
|
||||
%
|
||||
\begin{align}
|
||||
\prod_{a \tp A} \prod_{b \tp B} \prod_{q \tp A \equiv B} \var{coe}\ q\ a ≡ b →
|
||||
∏_{a \tp A} ∏_{b \tp B} ∏_{q \tp A ≡ B} \var{coe}\ q\ a ≡ b →
|
||||
\Path\ (λ \; i → q\ i)\ a\ b
|
||||
\end{align}
|
||||
%
|
||||
|
|
@ -1263,15 +1263,15 @@ 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
|
||||
p \tp & \Path\ (\lambda\; i → \Arrow\ X\ Y)\quad
|
||||
&& f\quad && y_𝒜 \x y_ℬ \\
|
||||
& \Path\ (\lambda\; i \to \Phi\ (p\ i))\quad
|
||||
& \Path\ (\lambda\; i → \Phi\ (p\ i))\quad
|
||||
&& \phi_f\quad && \phi_{y_𝒜 \x y_ℬ}
|
||||
\end{alignat}
|
||||
%
|
||||
Here $\Phi$ is given as:
|
||||
$$
|
||||
\prod_{f \tp \Arrow\ Y\ X}
|
||||
∏_{f \tp \Arrow\ Y\ X}
|
||||
( x_𝒜 \lll f ≡ y_𝒜 )
|
||||
×
|
||||
( x_ℬ \lll f ≡ y_ℬ )
|
||||
|
|
@ -1282,7 +1282,7 @@ again use the same trick we did in \ref{eq:propAreInversesGen} and prove this
|
|||
more general result:
|
||||
%
|
||||
$$
|
||||
\prod_{f \tp \Arrow\ Y\ X} \isProp\ (
|
||||
∏_{f \tp \Arrow\ Y\ X} \isProp\ (
|
||||
( x_𝒜 \lll f ≡ y_𝒜 )
|
||||
×
|
||||
( x_ℬ \lll f ≡ y_ℬ )
|
||||
|
|
@ -1298,7 +1298,7 @@ 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)
|
||||
∏_{A, B \tp \Object} \isProp\ (\var{Product}\ \bC\ A\ B)
|
||||
\end{align}
|
||||
%
|
||||
\section{Functors and natural transformations}
|
||||
|
|
@ -1331,7 +1331,7 @@ Given two functors between categories $\bC$ and $\bD$. Name them
|
|||
$\FunF$ and $\FunG$. A natural transformation is a family of arrows:
|
||||
%
|
||||
\begin{align*}
|
||||
\prod_{C \tp ℂ.\Object} \bD.\Arrow\ (\omapF\ C)\ (\omapG\ C)
|
||||
∏_{C \tp ℂ.\Object} \bD.\Arrow\ (\omapF\ C)\ (\omapG\ C)
|
||||
\end{align*}
|
||||
%
|
||||
This family of arrows can be seen as the data. If $\theta$ is a
|
||||
|
|
@ -1340,7 +1340,7 @@ $\theta$) at $C$. The laws of this family of morphism is the
|
|||
naturality condition:
|
||||
%
|
||||
\begin{align*}
|
||||
\prod_{f \tp ℂ.\Arrow\ A\ B}
|
||||
∏_{f \tp ℂ.\Arrow\ A\ B}
|
||||
(θ\ B) \dlll (\FunF.\fmap\ f) ≡ (\FunG.\fmap\ f) \dlll (θ\ A)
|
||||
\end{align*}
|
||||
%
|
||||
|
|
@ -1409,9 +1409,9 @@ The Kleisli-formulation consists of the following data:
|
|||
\begin{split}
|
||||
\label{eq:monad-kleisli-data}
|
||||
\omapR & \tp \Object → \Object \\
|
||||
\pure & \tp % \prod_{X \tp Object}
|
||||
\pure & \tp % ∏_{X \tp Object}
|
||||
\Arrow\ X\ (\omapR\ X) \\
|
||||
\bind & \tp % \prod_{X\;Y \tp Object} → \Arrow\ X\ (\omapR\ Y)
|
||||
\bind & \tp % ∏_{X\;Y \tp Object} → \Arrow\ X\ (\omapR\ Y)
|
||||
\Arrow\ (\omapR\ X)\ (\omapR\ Y)
|
||||
\end{split}
|
||||
\end{align}
|
||||
|
|
@ -1420,9 +1420,9 @@ The objects $X$ and $Y$ are implicitly universally quantified. With this data we
|
|||
%
|
||||
\begin{align*}
|
||||
\fish & \tp \Arrow\ A\ (\omapR\ B)
|
||||
\to \Arrow\ B\ (\omapR\ C)
|
||||
\to \Arrow\ A\ (\omapR\ C) \\
|
||||
f \fish g & \defeq f \rrr (\bind\ g)
|
||||
→ \Arrow\ B\ (\omapR\ C)
|
||||
→ \Arrow\ A\ (\omapR\ C) \\
|
||||
f \fish g & ≜ f \rrr (\bind\ g)
|
||||
\end{align*}
|
||||
%
|
||||
It is interesting to note here that this formulation does not talk about natural
|
||||
|
|
@ -1461,13 +1461,13 @@ In the monoidal formulation we can define $\bind$:
|
|||
\newcommand\pureX{\wideoverbar{\pure}}%
|
||||
\newcommand\fmapX{\wideoverbar{\fmap}}%
|
||||
\begin{align}
|
||||
\bind\ f \defeq \join \lll \fmap\ f
|
||||
\bind\ f ≜ \join \lll \fmap\ f
|
||||
\end{align}
|
||||
%
|
||||
And likewise in the Kleisli formulation we can define $\join$:
|
||||
%
|
||||
\begin{align}
|
||||
\join \defeq \bind\ \identity
|
||||
\join ≜ \bind\ \identity
|
||||
\end{align}
|
||||
%
|
||||
It now remains to show that this construction indeed gives rise to a
|
||||
|
|
@ -1483,9 +1483,9 @@ formulation we pick:
|
|||
%
|
||||
\begin{align}
|
||||
\begin{split}
|
||||
\omapR & \defeq \omapR \\
|
||||
\pure & \defeq \pure \\
|
||||
\bind\ f & \defeq \join \lll \fmap\ f
|
||||
\omapR & ≜ \omapR \\
|
||||
\pure & ≜ \pure \\
|
||||
\bind\ f & ≜ \join \lll \fmap\ f
|
||||
\end{split}
|
||||
\end{align}
|
||||
%
|
||||
|
|
@ -1549,9 +1549,9 @@ the monoidal formulation we pick:
|
|||
%
|
||||
\begin{align}
|
||||
\begin{split}
|
||||
\EndoR & \defeq (\omapR, \bind\ (\pure \lll f)) \\
|
||||
\pure & \defeq \pure \\
|
||||
\join & \defeq \bind\ \identity
|
||||
\EndoR & ≜ (\omapR, \bind\ (\pure \lll f)) \\
|
||||
\pure & ≜ \pure \\
|
||||
\join & ≜ \bind\ \identity
|
||||
\end{split}
|
||||
\end{align}
|
||||
%
|
||||
|
|
@ -1570,23 +1570,23 @@ that the two mappings sketched above are indeed inverses of each
|
|||
other. To recap, these maps are:
|
||||
%
|
||||
\begin{align*}
|
||||
\toKleisli & \tp \var{Kleisli} \to \var{Monoidal} \\
|
||||
\toKleisli & \defeq \lambda\ (\omapR, \pure, \bind)
|
||||
\to (\EndoR, \pure, \bind\ \identity)
|
||||
\toKleisli & \tp \var{Kleisli} → \var{Monoidal} \\
|
||||
\toKleisli & ≜ \lambda\ (\omapR, \pure, \bind)
|
||||
→ (\EndoR, \pure, \bind\ \identity)
|
||||
\end{align*}
|
||||
%
|
||||
Where $\EndoR \defeq (\omapR, \bind\ (\pure \lll f))$. The proof that
|
||||
Where $\EndoR ≜ (\omapR, \bind\ (\pure \lll f))$. The proof that
|
||||
this is indeed a functor is left implicit as well as the monad laws.
|
||||
Likewise the proof that $\pure$ and $\bind\ \identity$ are natural
|
||||
transformations are left implicit. The inverse map will be:
|
||||
%
|
||||
\begin{align*}
|
||||
\toMonoidal & \tp \var{Monoidal} \to \var{Kleisli} \\
|
||||
\toMonoidal & \defeq \lambda\ (\EndoR, \pureNT, \joinNT)
|
||||
\to (\omapR, \pure, \bind)
|
||||
\toMonoidal & \tp \var{Monoidal} → \var{Kleisli} \\
|
||||
\toMonoidal & ≜ \lambda\ (\EndoR, \pureNT, \joinNT)
|
||||
→ (\omapR, \pure, \bind)
|
||||
\end{align*}
|
||||
%
|
||||
Where $\bind\ f \defeq \join \lll \fmap\ f$. Again the monad laws are
|
||||
Where $\bind\ f ≜ \join \lll \fmap\ f$. Again the monad laws are
|
||||
left implicit. Now we must show:
|
||||
%
|
||||
\begin{align}
|
||||
|
|
|
|||
Loading…
Reference in New Issue