读者，作家单子

令$C$为CCC。设$(\times)$为的乘积双函子$C$。由于Cat是CCC，因此我们可以咖喱$(\times)$：

$curry (\times) : C \rightarrow(C \Rightarrow C)$

$curry (\times) A = \lambda B. A \times B$

函子范畴$C \Rightarrow C$具有通常monoidal结构。 所述的半群$C \Rightarrow C$处于单子$C$。 我们将有限积视为上的单调结构$C$。

$curry (\times) 1 \cong id$

$\forall A\ B. curry (\times) (A\times B) \cong (curry (\times) A) \circ (curry (\times) B)$

因此$(curry (\times))$保留了单曲面结构，因此将一个单面体传输到单子，而将共形子传输到共体。即，它将一个任意的等分面词传输$w$到$(Writer\ w)$ monad（请看定义— $w$必须是一个等分词）。类似地，它将对角线类彗星传输到Coreader comonad。

现在，为具体起见，我将介绍Writer的构造。

开始。实际上，它们简单地具有在Haskell不同的名称。我们有一个Haskell的幺⟨ 瓦特，中号一个p p ë Ñ d ，米ê 米p 吨ÿ ⟩：$Writer = Coreader = curry(\times)$ $\langle w, mappend, mempty\rangle$

$mappend : w\times w \to w$

$mempty : 1 \to w$

作家是一个仿函数，所以它也必须映射的态射，如和米Ë 米p 吨ÿ。我写如下，尽管在Haskell中无效：$mappend$$mempty$

$Writer\ mappend : Writer(w\times w) \to Writer\ w$

是自然变换，在一个态射 Ç ⇒ Ç。通过的性质 Ç ù - [R [R ý （× ）是一个函数，它接受一个∈ Ô b （C ^ ），并给出在一个态射 Ç：$Writer\ mappend$$C\Rightarrow C$$curry(\times)$$a \in Ob(C)$$C$

$Writer\ mappend\ a = mappend\times(id(a)) : Writer (w\times w) a \to Writer\ w\ a$

非正式地，类型的总和分量瓦特和泵一个完好。这正是Haskell中Writer的定义。一个障碍是，单子⟨ w ^ [R 我牛逼è [R w ^ ，μ ，η ⟩我们需要$Writer\ mappend\ a$$w$$a$$\langle Writer\ w,\mu,\eta\rangle$

$\mu : Writer\ w \circ Writer\ w \to Writer\ w$

即类型不兼容。但是，这些函子是同构的：由通常的关联器用于有限的产品，其是一种天然的同构＆cong; λ 一个。瓦特× （瓦特× 一）= w ^ - [R 我吨Ë ř 瓦特∘ W¯¯ ř 我吨Ë ř $Writer (w\times w) = \lambda a. (w\times w)\times a$$\cong \lambda a. w\times(w\times a) = Writer\ w \circ Writer\ w$。然后，我们定义经由W¯¯ ř 我吨Ë ř 米一个p p ê Ñ d。我省略的构造的η经由米Ë 米p 吨ÿ。$\mu$$Writer\ mappend$$\eta$$mempty$

作家，作为一个算符，保存交换图表，即，蜜饯幺半等式，所以我们有理所当然证明等式 =幺半群中（Ç ⇒ C ^ ） =一个单子在Ç。结束。$\langle Writer\ w,\mu,\eta\rangle$$(C\Rightarrow C)$$C$

怎么样阅读器和Cowriter？如Coreader的定义中所述，Reader与Coreader相连，请参见上面的链接。同样，Cowriter与Writer相邻。我没有找到Cowriter的定义，所以我通过表中所示的类比来弥补：

{- base, Hackage.category-extras -}
import Control.Comonad
import Data.Monoid
data Cowriter w a = Cowriter (w -> a)
instance Functor (Cowriter w) where
    fmap f (Cowriter g) = Cowriter (f . g)
instance Monoid w => Copointed (Cowriter w) where
    extract (Cowriter g) = g mempty
instance Monoid w => Comonad (Cowriter w) where
    duplicate (Cowriter g) = Cowriter
        (\w' -> Cowriter (\w -> g (w `mappend` w')))

以下是这些（共）单原子的简化定义。fr_ob F表示仿函数F在对象上的映射，fr_mor F表示仿函数F在射态上的映射。有一个幺对象在Ç。$\langle w,\hat{+},\hat{0}\rangle$$C$

• 作家
  • $fr\_ob (Writer\ w) a = a\times w$
  • $fr\_mor (Writer\ w) f = \lambda \langle a_0,w_2\rangle. \langle a_0,f\ w_2\rangle$
  • $\eta a = \lambda a_0. \langle a_0, \hat{0} \rangle$
  • $\mu a = \lambda \langle \langle a_0,w_1\rangle,w_0\rangle. \langle a_0, w_0\hat{+}w_1\rangle$
• 读者
  • $fr\_ob (Reader\ r) a = r\to a$
  • $fr\_mor (Reader\ r) f = \lambda g\ r_0. f (g\ r_0)$
  • $\eta a = \lambda a_0\ r_0. a_0$
  • $\mu a = \lambda f\ r_0. f\ r_0\ r_0$
• 共同阅读器
  • $fr\_ob (Coreader\ r) a = r\times a$
  • $fr\_mor (Coreader\ r) f = \lambda \langle r_0,a_0\rangle. \langle f\ r_0,a_0\rangle$
  • $\eta a = \lambda \langle r_0,a_0\rangle. a_0$
  • $\mu a = \lambda \langle r_0,a_0\rangle. \langle r_0,\langle r_0,a_0\rangle\rangle$
• Cowriter
  • $fr\_ob (Cowriter\ w) a = w\to a$
  • $fr\_mor (Cowriter\ w) f = \lambda g\ r_0. f (g\ r_0)$
  • $\eta a = \lambda f. f\ \hat{0}$
  • $\mu a = \lambda f\ w_1 w_0. f (w_0\hat{+}w_1)$

The question is that the adjunction in $C$ relates functors, not monads. I do not see how the adjunction implies "Coreader is a comonad" $\to$ "Reader is a monad" and "Writer is a monad" $\to$ "Cowriter is a comonad".

Remark. I am struggling to provide more context. It requires some work. Especially, if you require categorical purity and those (co)monads were introduced for programmers. Keep nagging! ;)

Yes, if a monad $M : C \to C$ has a right adjoint $N$, then $N$ automatically inherits a comonad structure.

The general category-theoretic setting to understand this is as follows. Let $C$ and $D$ be two categories. Write $\mathrm{Fun}(C, D)$ for the categeory of functors from $C$ to $D$; Its objects are functors and its morphisms natural transformations. Write $\mathrm{Fun}^L(C, D)$ for the full subcategory of $\mathrm{Fun}(C, D)$ on the functors which have right adjoints (in other words, we consider functors $C \to D$ with right adjoints and arbitrary natural transformations between them). Write $F^R : D \to C$ for the right adjoint of a functor $F : C \to D$. Then $-^R : \mathrm{Fun}^L(C, D) \to \mathrm{Fun}(D, C)$ is a contravariant functor: if $\alpha : F \to G$ is a natural transformation then there is an induced natural transformation $\alpha^R : G^R \to F^R$.

If $C = D$, then $\mathrm{Fun}(C, C)$ has a monoidal structure given by composition and so does $\mathrm{Fun}^L(C, D)$, because the composition of left adjoints is a left adjoint. Specifically, $(FG)^R = G^R F^R$, so $-^R$ is an antimonoidal contravariant functor. If you apply $-^R$ to the structural natural transformations which equip a functor $M$ with the structure of a monad, what you get out is a comonad.

By the way, this:

Let $(\times)$ be a product bifunctor in $C$. As $C$ is CCC, we can curry $(\times)$

is slightly incorrect. For one, the usual terminology would be (if I'm not mistaken) that $\times$ is a bifunctor over or on $C$. "In" typically means constructions using the arrows and objects of a category, whereas functors "on" categories refer to constructions relating multiple categories. And the product bifunctor isn't a construction within a Cartesian category.

And this relates to the larger inaccuracy: the ability to curry the product bifunctor has nothing to do with $C$ being Cartesian closed. Rather, it is possible because $Cat$, the category of categories (insert caveats) is Cartesian closed. So the currying in question is given by:

where $C \times D$ is a product of categories, and $E^D$ is the category of functors $F : D \to E$. This works regardless of whether $C$, $D$ and $E$ are Cartesian closed, though. When we let $C = D = E$, we get:

But this is merely a special case of:

Consider the adjunction $\langle F,G,\epsilon,\eta \rangle$. For every such adjunction we have a monad $\langle GF, \eta, G\epsilon F\rangle$ and also a comonad $\langle FG, \epsilon, F\eta G\rangle$. Notably, $F$ and $G$ need not be endofunctors, and in general they aren't (e.g., the list monad is an adjuction between the free and forgetful functors between $\mathbf{Set}$ and $\mathbf{Mon}$).

So, what you want to do is take Reader (or Writer) and decompose it into the adjoint functors which give rise to the monad and the corresponding comonad. Which is to say that the connection between Reader and Coreader (or Writer and Cowriter) isn't the one you're looking for.

And it's probably better to think of currying as $-^\ast : \text{hom}({-}\times A, {=}) \cong \text{hom}({-}, {=}^A)$, i.e. $\forall X,Y.~ \lbrace f : X\times A \to Y \rbrace \cong \lbrace f^\ast : X \to Y^A \rbrace$. Or if it helps, $-^\ast : \text{hom}({-}\times A, {=}\times 1) \cong \text{hom}({-}^1, {=}^A)$