update to use mathsf
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@ -32,7 +32,7 @@ open Σ
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With just that notation, the problem may not be clear.
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$2$ represents the set with only two elements in it, which is the booleans. The
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elements of $2$ are $\textrm{true}$ and $\textrm{false}$.
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elements of $2$ are $\mathsf{true}$ and $\mathsf{false}$.
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```
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data 𝟚 : Set where
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@ -53,8 +53,8 @@ shown by Theorem 4.1.3 in the [HoTT book][book].
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[1]: https://en.wikipedia.org/wiki/Homotopy#Homotopy_equivalence
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- there exists a $g : B \rightarrow A$
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- $f \circ g$ is homotopic to the identity map $\textrm{id}_B : B \rightarrow B$
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- $g \circ f$ is homotopic to the identity map $\textrm{id}_A : A \rightarrow A$
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- $f \circ g$ is homotopic to the identity map $\mathsf{id}_B : B \rightarrow B$
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- $g \circ f$ is homotopic to the identity map $\mathsf{id}_A : A \rightarrow A$
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We can write this in Agda like this:
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@ -92,7 +92,7 @@ function along with proof of its equivalence properties using a dependent
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pair[^dependent-pair]:
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[^dependent-pair]: A dependent pair (or $\Sigma$-type) is like a regular pair $\langle x, y\rangle$, but where $y$ can depend on $x$.
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For example, $\langle x , \textrm{isPrime}(x) \rangle$.
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For example, $\langle x , \mathsf{isPrime}(x) \rangle$.
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In this case it's useful since we can carry the equivalence information along with the function itself.
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This type is rather core to Martin-Löf Type Theory, you can read more about it [here][dependent-pair].
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@ -184,8 +184,8 @@ Now we need another function in the other direction. We can't case-split on
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functions, but we can certainly case-split on their output. Specifically, we can
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differentiate `id` from `neg` by their behavior when being called on `true`:
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- $\textrm{id}(\textrm{true}) :\equiv \textrm{true}$
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- $\textrm{neg}(\textrm{true}) :\equiv \textrm{false}$
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- $\mathsf{id}(\mathsf{true}) :\equiv \mathsf{true}$
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- $\mathsf{neg}(\mathsf{true}) :\equiv \mathsf{false}$
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```
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g : (𝟚 ≃ 𝟚) → 𝟚
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@ -207,7 +207,7 @@ everything to true, since it can't possibly have an inverse.
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We'll come back to this later.
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First, let's show that $g \circ f \sim \textrm{id}$. This one is easy, we can
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First, let's show that $g \circ f \sim \mathsf{id}$. This one is easy, we can
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just case-split. Each of the cases reduces to something that is definitionally
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equal, so we can use `refl`.
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@ -217,7 +217,7 @@ g∘f true = refl
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g∘f false = refl
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```
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Now comes the complicated case: proving $f \circ g \sim \textrm{id}$.
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Now comes the complicated case: proving $f \circ g \sim \mathsf{id}$.
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> [!admonition: NOTE]
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> Since Agda's comment syntax is `--`, the horizontal lines in the code below
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@ -235,9 +235,9 @@ module f∘g-case where
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f∘g eqv = goal eqv
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```
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Now our goal is to show that for any equivalence $\textrm{eqv} : 2 \simeq 2$,
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Now our goal is to show that for any equivalence $\mathsf{eqv} : 2 \simeq 2$,
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applying $f ∘ g$ to it is the same as not doing anything. We can evaluate the
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$g(\textrm{eqv})$ a little bit to give us a more detailed goal:
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$g(\mathsf{eqv})$ a little bit to give us a more detailed goal:
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```
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goal2 :
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@ -273,27 +273,27 @@ a syntax known as [with-abstraction]:
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We can now case-split on $b$, which is the output of calling $f$ on the
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equivalence returned by $g$. This means that for the `true` case, we need to
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show that $f(b) = \textrm{bool-eqv}$ (which is based on `id`) is equivalent to
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show that $f(b) = \mathsf{booleqv}$ (which is based on `id`) is equivalent to
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the equivalence that generated the `true`.
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Let's start with the `id` case; we just need to show that for every equivalence
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$e$ where running the equivalence function on `true` also returned `true`, $e
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\equiv f(\textrm{true})$.
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\equiv f(\mathsf{true})$.
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Unfortunately, we don't know if this is true unless our equivalences are _mere
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propositions_, meaning if two functions are identical, then so are their
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equivalences.
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$$
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\textrm{isProp}(P) :\equiv \prod_{x, y: P}(x \equiv y)
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\mathsf{isProp}(P) :\equiv \prod_{x, y \, : \, P}(x \equiv y)
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$$
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<small>Definition 3.3.1 from the [HoTT book][book]</small>
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Applying this to $\textrm{isEquiv}(f)$, we get the property:
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Applying this to $\mathsf{isEquiv}(f)$, we get the property:
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$$
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\sum_{f : A → B} \left( \prod_{e_1, e_2 : \textrm{isEquiv}(f)} e_1 \equiv e_2 \right)
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\sum_{f : A → B} \left( \prod_{e_1, e_2 \, : \, \mathsf{isEquiv}(f)} e_1 \equiv e_2 \right)
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$$
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This proof is shown later in the book, so I will use it here directly without proof[^equiv-isProp]:
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@ -314,7 +314,7 @@ equivalences must not map both values to a single one.
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This way, we can pin the behavior of the function on all inputs by just using
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its behavior on `true`, since its output on `false` must be _different_.
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We can use a proof that [$\textrm{true} \not\equiv \textrm{false}$][true-not-false] that I've shown previously.
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We can use a proof that [$\mathsf{true} \not\equiv \mathsf{false}$][true-not-false] that I've shown previously.
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[true-not-false]: https://mzhang.io/posts/proving-true-from-false/
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@ -63,7 +63,7 @@ const excerpt = remarkPluginFrontmatter.excerpt?.replaceAll("\n", "");
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<span class="tags">
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{
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post.data.draft && (
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<a href="/drafts" class="tag draft">
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<a href="/drafts/" class="tag draft">
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<i class="fa fa-warning" aria-hidden="true" />
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<span class="text">draft</span>
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</a>
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