What are the domains of the multiplication and unit morphisms of a monoid object?The abstract definition of commutative monoidsRooted trees morphisms and categoriesDefining a monoidal category without elementsMonoidal product is coproduct in category of commutative monoidsTaking the quotient of a free monoidal category modulo a relationTrying to understand significance of monoid as a one object categoryMonoid as a single object categoryTrying to understand monoids in Set as general monoid in monoidal categoryA monoid is $(M, •)$ or $(M, μ, η)$?Definition of a monoid object
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What are the domains of the multiplication and unit morphisms of a monoid object?
The abstract definition of commutative monoidsRooted trees morphisms and categoriesDefining a monoidal category without elementsMonoidal product is coproduct in category of commutative monoidsTaking the quotient of a free monoidal category modulo a relationTrying to understand significance of monoid as a one object categoryMonoid as a single object categoryTrying to understand monoids in Set as general monoid in monoidal categoryA monoid is $(M, •)$ or $(M, μ, η)$?Definition of a monoid object
$begingroup$
I'm trying to understand what a Monoid is from category theory perspective, but I'm a bit confused with notation used to describe it. Here is Wikipedia:
In category theory, a monoid (or monoid object) $(M, mu, eta)$ in a monoidal category $(mathcalC, otimes, I)$ is an object $M$ together with two morphisms
$mu: M otimes M to M$ called multiplication,
$eta: I to M$ called unit, [...]
My confusion is about morphism notation. Why is the binary operation $otimes$ a part of the morphism notation? My understanding of morphism is that it's a kind of function that can map one type to another (domain to codomain) like $M to M$... Why is the operation $otimes$ a part of the domain in the definition?
The second confusion is about $I$. Why $I$ is a domain... there is no $I$ object in a Monoid at all. It's just the neutral element of object $M$.
I understand that a Monoid is a category with one object, identity morphism and binary operation defined on this object, but the notation makes me think that I don't understand something.
Is $M otimes M$ somehow related to cartesian product, so the domain of the morphism is defined as $M times M$ ?
category-theory monoid monoidal-categories
New contributor
Bogdan Vakulenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I'm trying to understand what a Monoid is from category theory perspective, but I'm a bit confused with notation used to describe it. Here is Wikipedia:
In category theory, a monoid (or monoid object) $(M, mu, eta)$ in a monoidal category $(mathcalC, otimes, I)$ is an object $M$ together with two morphisms
$mu: M otimes M to M$ called multiplication,
$eta: I to M$ called unit, [...]
My confusion is about morphism notation. Why is the binary operation $otimes$ a part of the morphism notation? My understanding of morphism is that it's a kind of function that can map one type to another (domain to codomain) like $M to M$... Why is the operation $otimes$ a part of the domain in the definition?
The second confusion is about $I$. Why $I$ is a domain... there is no $I$ object in a Monoid at all. It's just the neutral element of object $M$.
I understand that a Monoid is a category with one object, identity morphism and binary operation defined on this object, but the notation makes me think that I don't understand something.
Is $M otimes M$ somehow related to cartesian product, so the domain of the morphism is defined as $M times M$ ?
category-theory monoid monoidal-categories
New contributor
Bogdan Vakulenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
5
$begingroup$
You should focus on understanding what a monoidal category is first, that seems to be your main problem.
$endgroup$
– Captain Lama
4 hours ago
add a comment |
$begingroup$
I'm trying to understand what a Monoid is from category theory perspective, but I'm a bit confused with notation used to describe it. Here is Wikipedia:
In category theory, a monoid (or monoid object) $(M, mu, eta)$ in a monoidal category $(mathcalC, otimes, I)$ is an object $M$ together with two morphisms
$mu: M otimes M to M$ called multiplication,
$eta: I to M$ called unit, [...]
My confusion is about morphism notation. Why is the binary operation $otimes$ a part of the morphism notation? My understanding of morphism is that it's a kind of function that can map one type to another (domain to codomain) like $M to M$... Why is the operation $otimes$ a part of the domain in the definition?
The second confusion is about $I$. Why $I$ is a domain... there is no $I$ object in a Monoid at all. It's just the neutral element of object $M$.
I understand that a Monoid is a category with one object, identity morphism and binary operation defined on this object, but the notation makes me think that I don't understand something.
Is $M otimes M$ somehow related to cartesian product, so the domain of the morphism is defined as $M times M$ ?
category-theory monoid monoidal-categories
New contributor
Bogdan Vakulenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm trying to understand what a Monoid is from category theory perspective, but I'm a bit confused with notation used to describe it. Here is Wikipedia:
In category theory, a monoid (or monoid object) $(M, mu, eta)$ in a monoidal category $(mathcalC, otimes, I)$ is an object $M$ together with two morphisms
$mu: M otimes M to M$ called multiplication,
$eta: I to M$ called unit, [...]
My confusion is about morphism notation. Why is the binary operation $otimes$ a part of the morphism notation? My understanding of morphism is that it's a kind of function that can map one type to another (domain to codomain) like $M to M$... Why is the operation $otimes$ a part of the domain in the definition?
The second confusion is about $I$. Why $I$ is a domain... there is no $I$ object in a Monoid at all. It's just the neutral element of object $M$.
I understand that a Monoid is a category with one object, identity morphism and binary operation defined on this object, but the notation makes me think that I don't understand something.
Is $M otimes M$ somehow related to cartesian product, so the domain of the morphism is defined as $M times M$ ?
category-theory monoid monoidal-categories
category-theory monoid monoidal-categories
New contributor
Bogdan Vakulenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Bogdan Vakulenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Arnaud D.
16.5k52445
16.5k52445
New contributor
Bogdan Vakulenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Bogdan VakulenkoBogdan Vakulenko
1083
1083
New contributor
Bogdan Vakulenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Bogdan Vakulenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
$begingroup$
You should focus on understanding what a monoidal category is first, that seems to be your main problem.
$endgroup$
– Captain Lama
4 hours ago
add a comment |
5
$begingroup$
You should focus on understanding what a monoidal category is first, that seems to be your main problem.
$endgroup$
– Captain Lama
4 hours ago
5
5
$begingroup$
You should focus on understanding what a monoidal category is first, that seems to be your main problem.
$endgroup$
– Captain Lama
4 hours ago
$begingroup$
You should focus on understanding what a monoidal category is first, that seems to be your main problem.
$endgroup$
– Captain Lama
4 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
An ordinary monoid is a set $M$ equipped with a map $mucolon Mtimes Mto M$ and an element $ein M$ such that certain axioms are satisfied.
Now the goal of the definition you quoted (monoid object in a monoidal category) is to generalize the notion of ordinary monoid to the most general categorical context possible.
The first step is to observe that in a general category it doesn't make any sense to talk about an "element" of an object. So let's replace the element $ein M$ with a map $ecolon 1to M$, where $1 = *$ is a set with a single element. The point is that (in the category of sets) specifying a map $1to M$ is exactly the same as specifying an element of $M$: given an element $ein M$, we can define a map $1to M$ by $*mapsto e$, and given a map $ecolon 1to M$, we can define an element of $M$ by $e(*)$.
Ok, now we want to generalize away from the category of sets and talk about monoids living in other categories. So let $mathcalC$ be a category. A monoid in $mathcalC$ should be an object $M$ in $mathcalC$ together with maps $mu$ and $e$ satisfying certain axioms. To make sense of the maps $mu$ and $e$, we need:
- A specified way of putting two copies of $M$ together to make the domain of $mu$, analogous to $Mtimes M$ in the case of sets. Let's denote this by $Motimes M$.
- A special object to be the domain of $e$, analogous to $1$ in the case of sets. Let's denote this by $I$.
The operation $otimes$ on objects and the special object $I$ are the basic data of a monoidal category.
If $mathcalC$ has finite products, you're welcome to define $otimes$ to be the product $times$ and $I$ to be the terminal object $1$. This is called the Cartesian monoidal structure. But the point is that we don't need $Motimes M$ to be the product and $I$ to be the terminal object in order to make sense of the definition of monoid.
We do need to assume a little bit more about $otimes$ and $I$ in order to make sense of the monoid axioms. In particular, we need $otimes$ to be associative (up to isomorphism), $$Motimes(Motimes M) cong (Motimes M)otimes M,$$ and we need $I$ to act as an identity for $otimes$ up to isomorphism, i.e. $$Mcong Iotimes Mcong Motimes I.$$
Making all of this appropriately functorial and natural leads directly to the full definition of monoidal category.
$endgroup$
$begingroup$
that was really helpful. thanks a lot.
$endgroup$
– Bogdan Vakulenko
3 hours ago
add a comment |
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1 Answer
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votes
$begingroup$
An ordinary monoid is a set $M$ equipped with a map $mucolon Mtimes Mto M$ and an element $ein M$ such that certain axioms are satisfied.
Now the goal of the definition you quoted (monoid object in a monoidal category) is to generalize the notion of ordinary monoid to the most general categorical context possible.
The first step is to observe that in a general category it doesn't make any sense to talk about an "element" of an object. So let's replace the element $ein M$ with a map $ecolon 1to M$, where $1 = *$ is a set with a single element. The point is that (in the category of sets) specifying a map $1to M$ is exactly the same as specifying an element of $M$: given an element $ein M$, we can define a map $1to M$ by $*mapsto e$, and given a map $ecolon 1to M$, we can define an element of $M$ by $e(*)$.
Ok, now we want to generalize away from the category of sets and talk about monoids living in other categories. So let $mathcalC$ be a category. A monoid in $mathcalC$ should be an object $M$ in $mathcalC$ together with maps $mu$ and $e$ satisfying certain axioms. To make sense of the maps $mu$ and $e$, we need:
- A specified way of putting two copies of $M$ together to make the domain of $mu$, analogous to $Mtimes M$ in the case of sets. Let's denote this by $Motimes M$.
- A special object to be the domain of $e$, analogous to $1$ in the case of sets. Let's denote this by $I$.
The operation $otimes$ on objects and the special object $I$ are the basic data of a monoidal category.
If $mathcalC$ has finite products, you're welcome to define $otimes$ to be the product $times$ and $I$ to be the terminal object $1$. This is called the Cartesian monoidal structure. But the point is that we don't need $Motimes M$ to be the product and $I$ to be the terminal object in order to make sense of the definition of monoid.
We do need to assume a little bit more about $otimes$ and $I$ in order to make sense of the monoid axioms. In particular, we need $otimes$ to be associative (up to isomorphism), $$Motimes(Motimes M) cong (Motimes M)otimes M,$$ and we need $I$ to act as an identity for $otimes$ up to isomorphism, i.e. $$Mcong Iotimes Mcong Motimes I.$$
Making all of this appropriately functorial and natural leads directly to the full definition of monoidal category.
$endgroup$
$begingroup$
that was really helpful. thanks a lot.
$endgroup$
– Bogdan Vakulenko
3 hours ago
add a comment |
$begingroup$
An ordinary monoid is a set $M$ equipped with a map $mucolon Mtimes Mto M$ and an element $ein M$ such that certain axioms are satisfied.
Now the goal of the definition you quoted (monoid object in a monoidal category) is to generalize the notion of ordinary monoid to the most general categorical context possible.
The first step is to observe that in a general category it doesn't make any sense to talk about an "element" of an object. So let's replace the element $ein M$ with a map $ecolon 1to M$, where $1 = *$ is a set with a single element. The point is that (in the category of sets) specifying a map $1to M$ is exactly the same as specifying an element of $M$: given an element $ein M$, we can define a map $1to M$ by $*mapsto e$, and given a map $ecolon 1to M$, we can define an element of $M$ by $e(*)$.
Ok, now we want to generalize away from the category of sets and talk about monoids living in other categories. So let $mathcalC$ be a category. A monoid in $mathcalC$ should be an object $M$ in $mathcalC$ together with maps $mu$ and $e$ satisfying certain axioms. To make sense of the maps $mu$ and $e$, we need:
- A specified way of putting two copies of $M$ together to make the domain of $mu$, analogous to $Mtimes M$ in the case of sets. Let's denote this by $Motimes M$.
- A special object to be the domain of $e$, analogous to $1$ in the case of sets. Let's denote this by $I$.
The operation $otimes$ on objects and the special object $I$ are the basic data of a monoidal category.
If $mathcalC$ has finite products, you're welcome to define $otimes$ to be the product $times$ and $I$ to be the terminal object $1$. This is called the Cartesian monoidal structure. But the point is that we don't need $Motimes M$ to be the product and $I$ to be the terminal object in order to make sense of the definition of monoid.
We do need to assume a little bit more about $otimes$ and $I$ in order to make sense of the monoid axioms. In particular, we need $otimes$ to be associative (up to isomorphism), $$Motimes(Motimes M) cong (Motimes M)otimes M,$$ and we need $I$ to act as an identity for $otimes$ up to isomorphism, i.e. $$Mcong Iotimes Mcong Motimes I.$$
Making all of this appropriately functorial and natural leads directly to the full definition of monoidal category.
$endgroup$
$begingroup$
that was really helpful. thanks a lot.
$endgroup$
– Bogdan Vakulenko
3 hours ago
add a comment |
$begingroup$
An ordinary monoid is a set $M$ equipped with a map $mucolon Mtimes Mto M$ and an element $ein M$ such that certain axioms are satisfied.
Now the goal of the definition you quoted (monoid object in a monoidal category) is to generalize the notion of ordinary monoid to the most general categorical context possible.
The first step is to observe that in a general category it doesn't make any sense to talk about an "element" of an object. So let's replace the element $ein M$ with a map $ecolon 1to M$, where $1 = *$ is a set with a single element. The point is that (in the category of sets) specifying a map $1to M$ is exactly the same as specifying an element of $M$: given an element $ein M$, we can define a map $1to M$ by $*mapsto e$, and given a map $ecolon 1to M$, we can define an element of $M$ by $e(*)$.
Ok, now we want to generalize away from the category of sets and talk about monoids living in other categories. So let $mathcalC$ be a category. A monoid in $mathcalC$ should be an object $M$ in $mathcalC$ together with maps $mu$ and $e$ satisfying certain axioms. To make sense of the maps $mu$ and $e$, we need:
- A specified way of putting two copies of $M$ together to make the domain of $mu$, analogous to $Mtimes M$ in the case of sets. Let's denote this by $Motimes M$.
- A special object to be the domain of $e$, analogous to $1$ in the case of sets. Let's denote this by $I$.
The operation $otimes$ on objects and the special object $I$ are the basic data of a monoidal category.
If $mathcalC$ has finite products, you're welcome to define $otimes$ to be the product $times$ and $I$ to be the terminal object $1$. This is called the Cartesian monoidal structure. But the point is that we don't need $Motimes M$ to be the product and $I$ to be the terminal object in order to make sense of the definition of monoid.
We do need to assume a little bit more about $otimes$ and $I$ in order to make sense of the monoid axioms. In particular, we need $otimes$ to be associative (up to isomorphism), $$Motimes(Motimes M) cong (Motimes M)otimes M,$$ and we need $I$ to act as an identity for $otimes$ up to isomorphism, i.e. $$Mcong Iotimes Mcong Motimes I.$$
Making all of this appropriately functorial and natural leads directly to the full definition of monoidal category.
$endgroup$
An ordinary monoid is a set $M$ equipped with a map $mucolon Mtimes Mto M$ and an element $ein M$ such that certain axioms are satisfied.
Now the goal of the definition you quoted (monoid object in a monoidal category) is to generalize the notion of ordinary monoid to the most general categorical context possible.
The first step is to observe that in a general category it doesn't make any sense to talk about an "element" of an object. So let's replace the element $ein M$ with a map $ecolon 1to M$, where $1 = *$ is a set with a single element. The point is that (in the category of sets) specifying a map $1to M$ is exactly the same as specifying an element of $M$: given an element $ein M$, we can define a map $1to M$ by $*mapsto e$, and given a map $ecolon 1to M$, we can define an element of $M$ by $e(*)$.
Ok, now we want to generalize away from the category of sets and talk about monoids living in other categories. So let $mathcalC$ be a category. A monoid in $mathcalC$ should be an object $M$ in $mathcalC$ together with maps $mu$ and $e$ satisfying certain axioms. To make sense of the maps $mu$ and $e$, we need:
- A specified way of putting two copies of $M$ together to make the domain of $mu$, analogous to $Mtimes M$ in the case of sets. Let's denote this by $Motimes M$.
- A special object to be the domain of $e$, analogous to $1$ in the case of sets. Let's denote this by $I$.
The operation $otimes$ on objects and the special object $I$ are the basic data of a monoidal category.
If $mathcalC$ has finite products, you're welcome to define $otimes$ to be the product $times$ and $I$ to be the terminal object $1$. This is called the Cartesian monoidal structure. But the point is that we don't need $Motimes M$ to be the product and $I$ to be the terminal object in order to make sense of the definition of monoid.
We do need to assume a little bit more about $otimes$ and $I$ in order to make sense of the monoid axioms. In particular, we need $otimes$ to be associative (up to isomorphism), $$Motimes(Motimes M) cong (Motimes M)otimes M,$$ and we need $I$ to act as an identity for $otimes$ up to isomorphism, i.e. $$Mcong Iotimes Mcong Motimes I.$$
Making all of this appropriately functorial and natural leads directly to the full definition of monoidal category.
edited 4 hours ago
answered 4 hours ago
Alex KruckmanAlex Kruckman
30.1k32758
30.1k32758
$begingroup$
that was really helpful. thanks a lot.
$endgroup$
– Bogdan Vakulenko
3 hours ago
add a comment |
$begingroup$
that was really helpful. thanks a lot.
$endgroup$
– Bogdan Vakulenko
3 hours ago
$begingroup$
that was really helpful. thanks a lot.
$endgroup$
– Bogdan Vakulenko
3 hours ago
$begingroup$
that was really helpful. thanks a lot.
$endgroup$
– Bogdan Vakulenko
3 hours ago
add a comment |
Bogdan Vakulenko is a new contributor. Be nice, and check out our Code of Conduct.
Bogdan Vakulenko is a new contributor. Be nice, and check out our Code of Conduct.
Bogdan Vakulenko is a new contributor. Be nice, and check out our Code of Conduct.
Bogdan Vakulenko is a new contributor. Be nice, and check out our Code of Conduct.
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5
$begingroup$
You should focus on understanding what a monoidal category is first, that seems to be your main problem.
$endgroup$
– Captain Lama
4 hours ago