fix typo in 3.14 (#340)

* fix typo in 3.14

* run prettier
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kkinugasa 2024-10-08 18:11:28 +09:00 committed by GitHub
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2 changed files with 4 additions and 4 deletions

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@ -33,8 +33,8 @@ feature-flags.
Afterwards, type `nix flake show` in the root directory of the project to see Afterwards, type `nix flake show` in the root directory of the project to see
all the available versions of this book. Then type `nix build .#<edition>` to all the available versions of this book. Then type `nix build .#<edition>` to
build the edition you want (Scala, OCaml, Reason and their printed build the edition you want (Scala, OCaml, Reason and their printed versions).
versions). For example, to build the Scala edition you'll have to type For example, to build the Scala edition you'll have to type
`nix build .#ctfp-scala`. For Haskell (the original version) that is just `nix build .#ctfp-scala`. For Haskell (the original version) that is just
`nix build .#ctfp`. `nix build .#ctfp`.

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@ -63,7 +63,7 @@ the roadmap:
Lawvere theory $\cat{L}$: an object in the category $\cat{Law}$. Lawvere theory $\cat{L}$: an object in the category $\cat{Law}$.
\item \item
Model $M$ of a Lawvere category: an object in the category\\ Model $M$ of a Lawvere category: an object in the category\\
$\cat{Mod}(\cat{Law}, \Set)$. $\cat{Mod}(\cat{L}, \Set)$.
\end{enumerate} \end{enumerate}
\begin{figure}[H] \begin{figure}[H]
@ -352,7 +352,7 @@ $n$. We can implement $F$ as the representable functor:
\[\cat{L}(n, -) \Colon \cat{L} \to \Set\] \[\cat{L}(n, -) \Colon \cat{L} \to \Set\]
To show that it's indeed free, all we have to do is to prove that it's a To show that it's indeed free, all we have to do is to prove that it's a
left adjoint to the forgetful functor: left adjoint to the forgetful functor:
\[\cat{Mod}(\cat{L}(n, -), M) \cong \Set(n, U(M))\] \[\cat{Mod}(\cat{L}, \Set)(\cat{L}(n, -), M) \cong \Set(n, U(M))\]
Let's simplify the right hand side: Let's simplify the right hand side:
\[\Set(n, U(M)) \cong \Set(n, M 1) \cong (M 1)^n \cong M n\] \[\Set(n, U(M)) \cong \Set(n, M 1) \cong (M 1)^n \cong M n\]
(I used the fact that a set of morphisms is isomorphic to the (I used the fact that a set of morphisms is isomorphic to the