Don't understand notation of morphisms in Monoid definitionWhat is Applicative Functor definition from the category theory POV?Examples of monoids/semigroups in programmingA monad is just a monoid in the category of endofunctors, what's the problem?“What part of Hindley-Milner do you not understand?”If “List” is a monoid, what is its “set”?A little category theoryWhat is the category-theoretical basis for the requirement that the Haskell “id” function must return the same value as passed in?Is my understanding of monoid valid?Why is `pure` only required for Applicative and not already for Functor?Free group monadHow is “a monoid on applicative functors” different than “a monoid in the category of endofunctors”?

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Don't understand notation of morphisms in Monoid definition


What is Applicative Functor definition from the category theory POV?Examples of monoids/semigroups in programmingA monad is just a monoid in the category of endofunctors, what's the problem?“What part of Hindley-Milner do you not understand?”If “List” is a monoid, what is its “set”?A little category theoryWhat is the category-theoretical basis for the requirement that the Haskell “id” function must return the same value as passed in?Is my understanding of monoid valid?Why is `pure` only required for Applicative and not already for Functor?Free group monadHow is “a monoid on applicative functors” different than “a monoid in the category of endofunctors”?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;








9















I'm trying to understand what Monoid is from a category theory perspective, but I'm a bit confused with the notation used to describe it. Here is Wikipedia:




In category theory, a monoid (or monoid object) (M, μ, η) in a monoidal category (C, ⊗, I) is an object M together with two morphisms



μ: M ⊗ M → M called multiplication,



η: I → M called unit




My confusion is about the morphism notation. Why is the binary operation a part of the morphism notation? My understanding of a morphism is that it's a kind of function that can map from one type to another (domain to codomain), like M → M. Why is the operation a part of the domain in the definition? The second confusion is about I. Why is I a domain? There is no I object in a Monoid at all. It's just a neutral element of the object M.



I understand that Monoid is a category with one object, an identity morphism, and a binary operation defined on this object, but the notation makes me think that I don't understand something.



Is M ⊗ M somehow related to the cartesian product, so that the domain of the morphism is defined as M x M?



Edit: I got a really helpful answer for my question on the Mathematics Stack Exchange.










share|improve this question



















  • 5





    Do you understand what monoid is from a set theory perspective? Definitions just are -- there's little answer that can be given to "why is the definition that way" other than "because we observe there are a bunch of things we want to talk about uniformly that are that way". So I can't really imagine answering that question sensibly. But I could imagine plausibly answering "How does this definition correspond with the various parts of the set theory definition?".

    – Daniel Wagner
    May 18 at 15:58












  • Okay. Since your (now deleted) comment says you grok the set theory definition, when I get back from lunch if there's no answer yet I'll write up a description of how the parts of the two definitions correspond; and the connection to the category with one object that's induced by the definition.

    – Daniel Wagner
    May 18 at 16:02











  • @daniel-wanger Yes, I understand what monoid is from set theory (at least the main idea). I even have some "non-stable" understanding of what it is in category theory. The main problem for me now (at least I think it's the main one)) is that I don't understand how to read notations. Is there kind of rule that say that on the left side (domain) there can be object only?. Because definition like this I -> M looks very strange for me. Like it's morphism from object I (which is not object) to M. Or M ⊗ M is kind of domain. But I would really appreciated for any answer.

    – Bogdan Vakulenko
    May 18 at 16:25






  • 8





    Warning! There are different but related notions at work! A category with one object corresponds to the traditional notion of a monoid (e.g., the ways you can get from where you are to where you are by hopping on the spot). The notion of monoidal category is something else: they can have much more interesting collections of objects in which (X) and I induce monoid-like structure (the way (,) on Haskell types is associative and absorbs () up to isomorphism).

    – pigworker
    May 18 at 16:33






  • 3





    The words "monoidal category" are essential, and it's where these operators you are asking about come from. You seem to be ignoring them.

    – luqui
    May 18 at 17:56


















9















I'm trying to understand what Monoid is from a category theory perspective, but I'm a bit confused with the notation used to describe it. Here is Wikipedia:




In category theory, a monoid (or monoid object) (M, μ, η) in a monoidal category (C, ⊗, I) is an object M together with two morphisms



μ: M ⊗ M → M called multiplication,



η: I → M called unit




My confusion is about the morphism notation. Why is the binary operation a part of the morphism notation? My understanding of a morphism is that it's a kind of function that can map from one type to another (domain to codomain), like M → M. Why is the operation a part of the domain in the definition? The second confusion is about I. Why is I a domain? There is no I object in a Monoid at all. It's just a neutral element of the object M.



I understand that Monoid is a category with one object, an identity morphism, and a binary operation defined on this object, but the notation makes me think that I don't understand something.



Is M ⊗ M somehow related to the cartesian product, so that the domain of the morphism is defined as M x M?



Edit: I got a really helpful answer for my question on the Mathematics Stack Exchange.










share|improve this question



















  • 5





    Do you understand what monoid is from a set theory perspective? Definitions just are -- there's little answer that can be given to "why is the definition that way" other than "because we observe there are a bunch of things we want to talk about uniformly that are that way". So I can't really imagine answering that question sensibly. But I could imagine plausibly answering "How does this definition correspond with the various parts of the set theory definition?".

    – Daniel Wagner
    May 18 at 15:58












  • Okay. Since your (now deleted) comment says you grok the set theory definition, when I get back from lunch if there's no answer yet I'll write up a description of how the parts of the two definitions correspond; and the connection to the category with one object that's induced by the definition.

    – Daniel Wagner
    May 18 at 16:02











  • @daniel-wanger Yes, I understand what monoid is from set theory (at least the main idea). I even have some "non-stable" understanding of what it is in category theory. The main problem for me now (at least I think it's the main one)) is that I don't understand how to read notations. Is there kind of rule that say that on the left side (domain) there can be object only?. Because definition like this I -> M looks very strange for me. Like it's morphism from object I (which is not object) to M. Or M ⊗ M is kind of domain. But I would really appreciated for any answer.

    – Bogdan Vakulenko
    May 18 at 16:25






  • 8





    Warning! There are different but related notions at work! A category with one object corresponds to the traditional notion of a monoid (e.g., the ways you can get from where you are to where you are by hopping on the spot). The notion of monoidal category is something else: they can have much more interesting collections of objects in which (X) and I induce monoid-like structure (the way (,) on Haskell types is associative and absorbs () up to isomorphism).

    – pigworker
    May 18 at 16:33






  • 3





    The words "monoidal category" are essential, and it's where these operators you are asking about come from. You seem to be ignoring them.

    – luqui
    May 18 at 17:56














9












9








9


3






I'm trying to understand what Monoid is from a category theory perspective, but I'm a bit confused with the notation used to describe it. Here is Wikipedia:




In category theory, a monoid (or monoid object) (M, μ, η) in a monoidal category (C, ⊗, I) is an object M together with two morphisms



μ: M ⊗ M → M called multiplication,



η: I → M called unit




My confusion is about the morphism notation. Why is the binary operation a part of the morphism notation? My understanding of a morphism is that it's a kind of function that can map from one type to another (domain to codomain), like M → M. Why is the operation a part of the domain in the definition? The second confusion is about I. Why is I a domain? There is no I object in a Monoid at all. It's just a neutral element of the object M.



I understand that Monoid is a category with one object, an identity morphism, and a binary operation defined on this object, but the notation makes me think that I don't understand something.



Is M ⊗ M somehow related to the cartesian product, so that the domain of the morphism is defined as M x M?



Edit: I got a really helpful answer for my question on the Mathematics Stack Exchange.










share|improve this question
















I'm trying to understand what Monoid is from a category theory perspective, but I'm a bit confused with the notation used to describe it. Here is Wikipedia:




In category theory, a monoid (or monoid object) (M, μ, η) in a monoidal category (C, ⊗, I) is an object M together with two morphisms



μ: M ⊗ M → M called multiplication,



η: I → M called unit




My confusion is about the morphism notation. Why is the binary operation a part of the morphism notation? My understanding of a morphism is that it's a kind of function that can map from one type to another (domain to codomain), like M → M. Why is the operation a part of the domain in the definition? The second confusion is about I. Why is I a domain? There is no I object in a Monoid at all. It's just a neutral element of the object M.



I understand that Monoid is a category with one object, an identity morphism, and a binary operation defined on this object, but the notation makes me think that I don't understand something.



Is M ⊗ M somehow related to the cartesian product, so that the domain of the morphism is defined as M x M?



Edit: I got a really helpful answer for my question on the Mathematics Stack Exchange.







haskell category-theory monoids






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited May 18 at 20:40









4castle

22.5k54475




22.5k54475










asked May 18 at 15:41









Bogdan VakulenkoBogdan Vakulenko

1,775218




1,775218







  • 5





    Do you understand what monoid is from a set theory perspective? Definitions just are -- there's little answer that can be given to "why is the definition that way" other than "because we observe there are a bunch of things we want to talk about uniformly that are that way". So I can't really imagine answering that question sensibly. But I could imagine plausibly answering "How does this definition correspond with the various parts of the set theory definition?".

    – Daniel Wagner
    May 18 at 15:58












  • Okay. Since your (now deleted) comment says you grok the set theory definition, when I get back from lunch if there's no answer yet I'll write up a description of how the parts of the two definitions correspond; and the connection to the category with one object that's induced by the definition.

    – Daniel Wagner
    May 18 at 16:02











  • @daniel-wanger Yes, I understand what monoid is from set theory (at least the main idea). I even have some "non-stable" understanding of what it is in category theory. The main problem for me now (at least I think it's the main one)) is that I don't understand how to read notations. Is there kind of rule that say that on the left side (domain) there can be object only?. Because definition like this I -> M looks very strange for me. Like it's morphism from object I (which is not object) to M. Or M ⊗ M is kind of domain. But I would really appreciated for any answer.

    – Bogdan Vakulenko
    May 18 at 16:25






  • 8





    Warning! There are different but related notions at work! A category with one object corresponds to the traditional notion of a monoid (e.g., the ways you can get from where you are to where you are by hopping on the spot). The notion of monoidal category is something else: they can have much more interesting collections of objects in which (X) and I induce monoid-like structure (the way (,) on Haskell types is associative and absorbs () up to isomorphism).

    – pigworker
    May 18 at 16:33






  • 3





    The words "monoidal category" are essential, and it's where these operators you are asking about come from. You seem to be ignoring them.

    – luqui
    May 18 at 17:56













  • 5





    Do you understand what monoid is from a set theory perspective? Definitions just are -- there's little answer that can be given to "why is the definition that way" other than "because we observe there are a bunch of things we want to talk about uniformly that are that way". So I can't really imagine answering that question sensibly. But I could imagine plausibly answering "How does this definition correspond with the various parts of the set theory definition?".

    – Daniel Wagner
    May 18 at 15:58












  • Okay. Since your (now deleted) comment says you grok the set theory definition, when I get back from lunch if there's no answer yet I'll write up a description of how the parts of the two definitions correspond; and the connection to the category with one object that's induced by the definition.

    – Daniel Wagner
    May 18 at 16:02











  • @daniel-wanger Yes, I understand what monoid is from set theory (at least the main idea). I even have some "non-stable" understanding of what it is in category theory. The main problem for me now (at least I think it's the main one)) is that I don't understand how to read notations. Is there kind of rule that say that on the left side (domain) there can be object only?. Because definition like this I -> M looks very strange for me. Like it's morphism from object I (which is not object) to M. Or M ⊗ M is kind of domain. But I would really appreciated for any answer.

    – Bogdan Vakulenko
    May 18 at 16:25






  • 8





    Warning! There are different but related notions at work! A category with one object corresponds to the traditional notion of a monoid (e.g., the ways you can get from where you are to where you are by hopping on the spot). The notion of monoidal category is something else: they can have much more interesting collections of objects in which (X) and I induce monoid-like structure (the way (,) on Haskell types is associative and absorbs () up to isomorphism).

    – pigworker
    May 18 at 16:33






  • 3





    The words "monoidal category" are essential, and it's where these operators you are asking about come from. You seem to be ignoring them.

    – luqui
    May 18 at 17:56








5




5





Do you understand what monoid is from a set theory perspective? Definitions just are -- there's little answer that can be given to "why is the definition that way" other than "because we observe there are a bunch of things we want to talk about uniformly that are that way". So I can't really imagine answering that question sensibly. But I could imagine plausibly answering "How does this definition correspond with the various parts of the set theory definition?".

– Daniel Wagner
May 18 at 15:58






Do you understand what monoid is from a set theory perspective? Definitions just are -- there's little answer that can be given to "why is the definition that way" other than "because we observe there are a bunch of things we want to talk about uniformly that are that way". So I can't really imagine answering that question sensibly. But I could imagine plausibly answering "How does this definition correspond with the various parts of the set theory definition?".

– Daniel Wagner
May 18 at 15:58














Okay. Since your (now deleted) comment says you grok the set theory definition, when I get back from lunch if there's no answer yet I'll write up a description of how the parts of the two definitions correspond; and the connection to the category with one object that's induced by the definition.

– Daniel Wagner
May 18 at 16:02





Okay. Since your (now deleted) comment says you grok the set theory definition, when I get back from lunch if there's no answer yet I'll write up a description of how the parts of the two definitions correspond; and the connection to the category with one object that's induced by the definition.

– Daniel Wagner
May 18 at 16:02













@daniel-wanger Yes, I understand what monoid is from set theory (at least the main idea). I even have some "non-stable" understanding of what it is in category theory. The main problem for me now (at least I think it's the main one)) is that I don't understand how to read notations. Is there kind of rule that say that on the left side (domain) there can be object only?. Because definition like this I -> M looks very strange for me. Like it's morphism from object I (which is not object) to M. Or M ⊗ M is kind of domain. But I would really appreciated for any answer.

– Bogdan Vakulenko
May 18 at 16:25





@daniel-wanger Yes, I understand what monoid is from set theory (at least the main idea). I even have some "non-stable" understanding of what it is in category theory. The main problem for me now (at least I think it's the main one)) is that I don't understand how to read notations. Is there kind of rule that say that on the left side (domain) there can be object only?. Because definition like this I -> M looks very strange for me. Like it's morphism from object I (which is not object) to M. Or M ⊗ M is kind of domain. But I would really appreciated for any answer.

– Bogdan Vakulenko
May 18 at 16:25




8




8





Warning! There are different but related notions at work! A category with one object corresponds to the traditional notion of a monoid (e.g., the ways you can get from where you are to where you are by hopping on the spot). The notion of monoidal category is something else: they can have much more interesting collections of objects in which (X) and I induce monoid-like structure (the way (,) on Haskell types is associative and absorbs () up to isomorphism).

– pigworker
May 18 at 16:33





Warning! There are different but related notions at work! A category with one object corresponds to the traditional notion of a monoid (e.g., the ways you can get from where you are to where you are by hopping on the spot). The notion of monoidal category is something else: they can have much more interesting collections of objects in which (X) and I induce monoid-like structure (the way (,) on Haskell types is associative and absorbs () up to isomorphism).

– pigworker
May 18 at 16:33




3




3





The words "monoidal category" are essential, and it's where these operators you are asking about come from. You seem to be ignoring them.

– luqui
May 18 at 17:56






The words "monoidal category" are essential, and it's where these operators you are asking about come from. You seem to be ignoring them.

– luqui
May 18 at 17:56













1 Answer
1






active

oldest

votes


















18















Is M ⊗ M some how related to cartesian product, so domain of the morphism is defined as M x M ?




Exactly. More specifically, we get those monoids that are expressed by the Monoid class from base by picking Hask (the category with all Haskell types as objects and all Haskell functions as morphisms) as C, (,) (the pair type constructor) as , and () (the unit type) as I. The signatures of μ and η, translated to Haskell, then become:



μ :: (M, M) -> M
η :: () -> M


By currying μ, and making use of how () -> M functions are in one-to-one correspondence to M values (all of them look like () -> m for some m), we get the familiar Monoid methods:



mappend :: M -> M -> M
mempty :: M


Note that the categorical definition is far more general than just Monoid. For instance, we might keep working in Hask while replacing (,) and () with their duals, Either and Void, thus getting:



μ :: Either A A -> A
η :: Void -> A


Every Haskell type is a monoid in this particular manner (μ is either id id, and η is absurd).




Another example is taking C to be the category of Haskell Functors (with natural transformations between them -- which I will write as type f ~> g = forall a. f a -> g a -- as morphisms), Compose as , and Identity as I:



-- Note the arrows here are ~>, and not ->
μ :: Compose M M ~> M
η :: Identity ~> M


These two are commonly written as:



-- "Inlining" the definitions of Compose, Identity, and ~>
join :: M (M a) -> M a
return :: a -> M a


In other words, a Monad is a monoid in the category of Functors (which is the Hask-specific version of "a monad is a monoid in the category of endofucntors"). It is worth mentioning that, as in the other example, this is not the only way to get monoids out of that category (see the final paragraph of this answer for pointers -- the rest of it, in fact, might be relevant reading, as it discusses the notion of monoidal category).






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    active

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    18















    Is M ⊗ M some how related to cartesian product, so domain of the morphism is defined as M x M ?




    Exactly. More specifically, we get those monoids that are expressed by the Monoid class from base by picking Hask (the category with all Haskell types as objects and all Haskell functions as morphisms) as C, (,) (the pair type constructor) as , and () (the unit type) as I. The signatures of μ and η, translated to Haskell, then become:



    μ :: (M, M) -> M
    η :: () -> M


    By currying μ, and making use of how () -> M functions are in one-to-one correspondence to M values (all of them look like () -> m for some m), we get the familiar Monoid methods:



    mappend :: M -> M -> M
    mempty :: M


    Note that the categorical definition is far more general than just Monoid. For instance, we might keep working in Hask while replacing (,) and () with their duals, Either and Void, thus getting:



    μ :: Either A A -> A
    η :: Void -> A


    Every Haskell type is a monoid in this particular manner (μ is either id id, and η is absurd).




    Another example is taking C to be the category of Haskell Functors (with natural transformations between them -- which I will write as type f ~> g = forall a. f a -> g a -- as morphisms), Compose as , and Identity as I:



    -- Note the arrows here are ~>, and not ->
    μ :: Compose M M ~> M
    η :: Identity ~> M


    These two are commonly written as:



    -- "Inlining" the definitions of Compose, Identity, and ~>
    join :: M (M a) -> M a
    return :: a -> M a


    In other words, a Monad is a monoid in the category of Functors (which is the Hask-specific version of "a monad is a monoid in the category of endofucntors"). It is worth mentioning that, as in the other example, this is not the only way to get monoids out of that category (see the final paragraph of this answer for pointers -- the rest of it, in fact, might be relevant reading, as it discusses the notion of monoidal category).






    share|improve this answer





























      18















      Is M ⊗ M some how related to cartesian product, so domain of the morphism is defined as M x M ?




      Exactly. More specifically, we get those monoids that are expressed by the Monoid class from base by picking Hask (the category with all Haskell types as objects and all Haskell functions as morphisms) as C, (,) (the pair type constructor) as , and () (the unit type) as I. The signatures of μ and η, translated to Haskell, then become:



      μ :: (M, M) -> M
      η :: () -> M


      By currying μ, and making use of how () -> M functions are in one-to-one correspondence to M values (all of them look like () -> m for some m), we get the familiar Monoid methods:



      mappend :: M -> M -> M
      mempty :: M


      Note that the categorical definition is far more general than just Monoid. For instance, we might keep working in Hask while replacing (,) and () with their duals, Either and Void, thus getting:



      μ :: Either A A -> A
      η :: Void -> A


      Every Haskell type is a monoid in this particular manner (μ is either id id, and η is absurd).




      Another example is taking C to be the category of Haskell Functors (with natural transformations between them -- which I will write as type f ~> g = forall a. f a -> g a -- as morphisms), Compose as , and Identity as I:



      -- Note the arrows here are ~>, and not ->
      μ :: Compose M M ~> M
      η :: Identity ~> M


      These two are commonly written as:



      -- "Inlining" the definitions of Compose, Identity, and ~>
      join :: M (M a) -> M a
      return :: a -> M a


      In other words, a Monad is a monoid in the category of Functors (which is the Hask-specific version of "a monad is a monoid in the category of endofucntors"). It is worth mentioning that, as in the other example, this is not the only way to get monoids out of that category (see the final paragraph of this answer for pointers -- the rest of it, in fact, might be relevant reading, as it discusses the notion of monoidal category).






      share|improve this answer



























        18












        18








        18








        Is M ⊗ M some how related to cartesian product, so domain of the morphism is defined as M x M ?




        Exactly. More specifically, we get those monoids that are expressed by the Monoid class from base by picking Hask (the category with all Haskell types as objects and all Haskell functions as morphisms) as C, (,) (the pair type constructor) as , and () (the unit type) as I. The signatures of μ and η, translated to Haskell, then become:



        μ :: (M, M) -> M
        η :: () -> M


        By currying μ, and making use of how () -> M functions are in one-to-one correspondence to M values (all of them look like () -> m for some m), we get the familiar Monoid methods:



        mappend :: M -> M -> M
        mempty :: M


        Note that the categorical definition is far more general than just Monoid. For instance, we might keep working in Hask while replacing (,) and () with their duals, Either and Void, thus getting:



        μ :: Either A A -> A
        η :: Void -> A


        Every Haskell type is a monoid in this particular manner (μ is either id id, and η is absurd).




        Another example is taking C to be the category of Haskell Functors (with natural transformations between them -- which I will write as type f ~> g = forall a. f a -> g a -- as morphisms), Compose as , and Identity as I:



        -- Note the arrows here are ~>, and not ->
        μ :: Compose M M ~> M
        η :: Identity ~> M


        These two are commonly written as:



        -- "Inlining" the definitions of Compose, Identity, and ~>
        join :: M (M a) -> M a
        return :: a -> M a


        In other words, a Monad is a monoid in the category of Functors (which is the Hask-specific version of "a monad is a monoid in the category of endofucntors"). It is worth mentioning that, as in the other example, this is not the only way to get monoids out of that category (see the final paragraph of this answer for pointers -- the rest of it, in fact, might be relevant reading, as it discusses the notion of monoidal category).






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        Is M ⊗ M some how related to cartesian product, so domain of the morphism is defined as M x M ?




        Exactly. More specifically, we get those monoids that are expressed by the Monoid class from base by picking Hask (the category with all Haskell types as objects and all Haskell functions as morphisms) as C, (,) (the pair type constructor) as , and () (the unit type) as I. The signatures of μ and η, translated to Haskell, then become:



        μ :: (M, M) -> M
        η :: () -> M


        By currying μ, and making use of how () -> M functions are in one-to-one correspondence to M values (all of them look like () -> m for some m), we get the familiar Monoid methods:



        mappend :: M -> M -> M
        mempty :: M


        Note that the categorical definition is far more general than just Monoid. For instance, we might keep working in Hask while replacing (,) and () with their duals, Either and Void, thus getting:



        μ :: Either A A -> A
        η :: Void -> A


        Every Haskell type is a monoid in this particular manner (μ is either id id, and η is absurd).




        Another example is taking C to be the category of Haskell Functors (with natural transformations between them -- which I will write as type f ~> g = forall a. f a -> g a -- as morphisms), Compose as , and Identity as I:



        -- Note the arrows here are ~>, and not ->
        μ :: Compose M M ~> M
        η :: Identity ~> M


        These two are commonly written as:



        -- "Inlining" the definitions of Compose, Identity, and ~>
        join :: M (M a) -> M a
        return :: a -> M a


        In other words, a Monad is a monoid in the category of Functors (which is the Hask-specific version of "a monad is a monoid in the category of endofucntors"). It is worth mentioning that, as in the other example, this is not the only way to get monoids out of that category (see the final paragraph of this answer for pointers -- the rest of it, in fact, might be relevant reading, as it discusses the notion of monoidal category).







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited May 18 at 17:40

























        answered May 18 at 16:24









        duplodeduplode

        25.1k45296




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