पाठक, लेखक सन्यासी

Question

बता दें कि $C$ एक CCC है । Let $(\times)$ पर एक उत्पाद बाइफंकेटर हो । जैसा कि कैट सीसीसी है, हम ( × ) करी कर सकते हैं :$C$$(\times)$

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

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

फ़नकार श्रेणी $C \Rightarrow C$ में सामान्य मोनोइडल संरचना होती है। में एक monoid $C \Rightarrow C$ में एक इकाई है $C$ । हम परिमित उत्पादों को पर मोनोइडल संरचना मानते हैं $C$।

$curry (\times) 1 \cong id$

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

इसलिए $(curry (\times))$ मोनोएडल संरचना को संरक्षित करता है, इसलिए यह एक मोनोड को एक मोनोड और एक कोमोनॉइड को कॉमनोड में स्थानांतरित करता है। अर्थात्, यह एक मनमाना monoid transports $w$ करने के लिए $(Writer\ w)$ इकाई (- परिभाषा पर नज़र $w$ एक monoid होना चाहिए)। इसी प्रकार यह विकर्ण कोमोनोइड को कोरियर कोमोनैड में स्थानांतरित करता है।

अब, संक्षिप्तता के लिए, मैं लेखक के निर्माण को प्रकट करता हूं।

शुरू। दरअसल $Writer = Coreader = curry(\times)$ , वे बस हास्केल में अलग-अलग नामों से हैं। हम एक है हास्केल monoid $\langle w, mappend, mempty\rangle$ :

$mappend : w\times w \to w$

$mempty : 1 \to w$

लेखक, एक functor है, इसलिए यह भी इस तरह के रूप में morphisms, नक्शा चाहिए और मीटर ई मीटर पी टी y । मैं इसे नीचे लिखता हूं, हालांकि यह हास्केल में अमान्य है:$mappend$$mempty$

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

एक प्राकृतिक परिवर्तन है, C ⇒ C में एक आकृतिवाद। तक के गुणों सी यू आर आर वाई ( × ) यह एक समारोह है, जो लेता है एक ∈ हे ख ( सी ) और में एक आकारिता देता है सी :$Writer\ mappend$$C\Rightarrow C$$curry(\times)$$a \in Ob(C)$$C$

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

अनौपचारिक रूप से, प्रकार की रकम घटकों डब्ल्यू और पंप एक अक्षुण्ण। हास्केल में यह बिल्कुल लेखक की परिभाषा है। एक बाधा यह है कि इकाई के लिए है ⟨ डब्ल्यू आर मैं टी ई आर डब्ल्यू , μ , η ⟩ हम जरूरत$Writer\ mappend\ a$$w$$a$$\langle Writer\ w,\mu,\eta\rangle$

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

अर्थात प्रकार की असंगति। लेकिन ये फंक्शनलर्स आइसोमॉर्फिक हैं: परिमित उत्पादों के लिए हमेशा की तरह Associator जो एक प्राकृतिक समाकृतिकता है द्वारा ≅ λ एक । डब्ल्यू × ( डब्ल्यू × एक ) = डब्ल्यू आर मैं टी ई आर डब्ल्यू ∘ डब्ल्यू आर मैं टी ई आर $Writer (w\times w) = \lambda a. (w\times w)\times a$$\cong \lambda a. w\times(w\times a) = Writer\ w \circ Writer\ w$। फिर हम परिभाषित के माध्यम से डब्ल्यू आर मैं टी ई आर एम एक पी पी ई एन डी । मैं का निर्माण छोड़ η के माध्यम से मीटर ई मीटर पी टी y ।$\mu$$Writer\ mappend$$\eta$$mempty$

लेखक, एक functor जा रहा है, विनिमेय आरेख, यानी बरकरार रखता है monoid समानताओं को बरकरार रखता है हम दी साबित कर दिया के लिए समानताओं के लिए है, इसलिए में = एक monoid ( सी ⇒ सी ) में = एक इकाई सी । समाप्त।$\langle Writer\ w,\mu,\eta\rangle$$(C\Rightarrow C)$$C$

रीडर और काउटर के बारे में क्या ? रीडर कोरीडर के निकट है, जैसा कि कॉरडर की परिभाषा में बताया गया है, ऊपर लिंक देखें। इसी तरह, काउटर राइटर के निकट है। मुझे कॉयराईटर की परिभाषा नहीं मिली, इसलिए मैंने इसे तालिका में दिखाए गए सादृश्य द्वारा बनाया है:

{- 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')))

नीचे उन (सह) भिक्षुओं की सरलीकृत परिभाषाएँ दी गई हैं। fr_ob F वस्तुओं पर एक फ़ंक्टर F की मैपिंग को दर्शाता है, fr_mor F, फ़ॉर्म्फ़र्स पर फ़ंफ़र्ट F की फ़ंडिंग की मैपिंग को दर्शाता है। वहाँ एक monoid वस्तु है में सी ।$\langle w,\hat{+},\hat{0}\rangle$$C$

• लेखक
  • $fr\_ob (Writer\ w) a = a\times w$
  • $fr\_mor (Writer\ w) f = \lambda \langle a_0,w_2\rangle. \langle a_0,f\ w_2\rangle$
  • $\eta a = \lambda a_0. \langle a_0, \hat{0} \rangle$
  • $\mu a = \lambda \langle \langle a_0,w_1\rangle,w_0\rangle. \langle a_0, w_0\hat{+}w_1\rangle$
• पाठक
  • $fr\_ob (Reader\ r) a = r\to a$
  • $fr\_mor (Reader\ r) f = \lambda g\ r_0. f (g\ r_0)$
  • $\eta a = \lambda a_0\ r_0. a_0$
  • $\mu a = \lambda f\ r_0. f\ r_0\ r_0$
• Coreader
  • $fr\_ob (Coreader\ r) a = r\times a$
  • $fr\_mor (Coreader\ r) f = \lambda \langle r_0,a_0\rangle. \langle f\ r_0,a_0\rangle$
  • $\eta a = \lambda \langle r_0,a_0\rangle. a_0$
  • $\mu a = \lambda \langle r_0,a_0\rangle. \langle r_0,\langle r_0,a_0\rangle\rangle$
• 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! ;)

Comments

Offer: You can take a screenshot of the table, and put the image here.

— M.S. Dousti

You should copy the question here.

— Dave Clarke

people downvoting should post a comment explaining why.

— Suresh Venkat

@Ohad. I confess that I introduced that change to try to provide the question with more context (as was originally found in the blog post originally referenced). I think beroal should spend more effort making his question self contained, for example, by defining what Reader and Writer and Coreader and Cowriter are in categorical terms or in Haskell or both, rather than assuming that we all know what is being referred to.

— Dave Clarke

@beroal: What I meant was that, as I don't use Haskell on a day to day basis, parsing the Haskell code and making the transition into CT is non-trivial for me, and perhaps others. By rephrasing the question in purely categorical terms, you are more likely to receive an answer quicker...

— Ohad Kammar

Answer 1

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.

Comments

And one should mention that some of these functors, for example ${-}^R$ is not really a functor but rather something like a pseudo-functor because it typically satisfies functoriality only up to canonical isomorphisms. Nevertheless, the main point is valid.

— Andrej Bauer

Answer 2

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:

$$Hom_{Cat}(C \times D, E) \cong Hom_{Cat}(C, E^D)$$

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:

$$\times : C \times C \to C$$ $$curry_\times : C \to C^C$$

But this is merely a special case of:

$$F : C \times D \to E$$ $$curry_F : C \to E^D$$

Comments

2 Dan Doel: Yes, yes, yes, thanks. I did the mistake while translating from the original post beroal.livejournal.com/23223.html .

— beroal

Answer 3

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, $-^\ast : \text{hom}({-}\times A, {=}\times 1) \cong \text{hom}({-}^1, {=}^A)$

Comments

2 wren ng thornton: I am not aware of any defining adjunction for Reader and Writer similar to adjunctions between Set and a category of algebraic structures. Or do you mean that every monad is defined by an adjunction as in "MacLane. Categories for the Working Mathematician. VI. Monads and Algebras. 2. Algebras for a Monad. Theorem 1 (Every monad is defined by its T-algebras)."? Can you be more specific? Actually my question is the conclusion of an attempt to define those (co)monads in elegant words as the list monad is.

— beroal

@beroal: I'm pretty sure Reader and Writer aren't adjoint, or at least I've yet to find a way to get the categories to work out for it. No, my point was that monads and comonads arise in "the same way", namely via an adjunction, as described above. I don't have a copy of MacLane, but yes $T$-algebras are the standard name for the trick above (but then again, all sorts of unrelated things are called "X-algebras", "Y-algebras",...).

— wren romano

Which description of the list monad are you trying to match the eloquence of? Given the free monoid functor $F:\mathbf{Set}\to\mathbf{Mon}$, the forgetful functor $U:\mathbf{Mon}\to\mathbf{Set}$, the unit transformation $\eta:\text{id}_{\mathbf{Set}}\to UF$, and the counit transformation $\epsilon:FU\to\text{id}_{\mathbf{Mon}}$ you have an adjunction $\langle F,U,\eta,\epsilon \rangle$. Which means you have a monad $\langle UF,\eta, U\epsilon F\rangle$, namely the list monad in $\mathbf{Set}$. And you get the list comonad in $\mathbf{Mon}$: $\langle FU,\epsilon, F\eta U\rangle$. Eloquent?

— wren romano

Functors (Reader a) and (Writer a) are adjoint, and that adjunction gives rise to the (State a) monad.

— beroal

"No, my point was that monads and comonads arise in "the same way", namely via an adjunction, as described above". If you get the monad and the comonad from the adjunction between categories Set and Mon, you get the monad on Set and the comonad on Mon — different categories. But Reader and Writer are on the same CCC category.

— beroal