It looks like you're new here. If you want to get involved, click one of these buttons!

- All Categories 2.4K
- Chat 505
- Study Groups 21
- Petri Nets 9
- Epidemiology 4
- Leaf Modeling 2
- Review Sections 9
- MIT 2020: Programming with Categories 51
- MIT 2020: Lectures 20
- MIT 2020: Exercises 25
- Baez ACT 2019: Online Course 339
- Baez ACT 2019: Lectures 79
- Baez ACT 2019: Exercises 149
- Baez ACT 2019: Chat 50
- UCR ACT Seminar 4
- General 75
- Azimuth Code Project 111
- Statistical methods 4
- Drafts 10
- Math Syntax Demos 15
- Wiki - Latest Changes 3
- Strategy 113
- Azimuth Project 1.1K
- - Spam 1
- News and Information 148
- Azimuth Blog 149
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 719

Options

*Monoids.*

A monoid \((M,∗,e) \) is

a set \(M \);

a function \(∗:M×M \to M \); and

an element \(e \in M \) called the identity;

subject to two laws:

Unit: the equations \(e∗m=m \) and \(m∗e=m \) hold for any \(m \in M \).

Associative: the equation \((m1∗m2)∗m3=m1∗(m2∗m3) \) holds for any \(m1,m2,m3 \in M \).

--

(a) Show that \((\mathbb{N},+,0) \) forms a monoid.

(b) A string in 0 and 1 is a (possibly) empty sequence of 0s and 1s; examples include 0, 11, 0110, 0101110 and so on. We write the empty string []. Let \(List_{0,1} \) be the set of strings in 0 and 1. Given two strings \(a \) and \(b \), we may concatenate them to form a new string \(ab \). Show that \(List_{0,1} \), together with concatenation and the empty string [], form a monoid.

(c) Explain why (prove that) every monoid can be viewed as a category with a single object.

## Comments

No takers?

`No takers?`

a:

Unit holds because \(0 + m = m \) and \(m + 0 = m \),

associativity holds because \((m_1 + m_2) + m_3 = m_1 + (m_2 + m_3) \)

b:

Unit holds because \([] ++ m = m \) and \( m ++ [] = m \)

associativity holds because \((m_1 ++ m_2) ++ m_3 = m_1 ++ (m_2 ++ m_3) \)

c:

A monoid can be viewed as a category with a single object (call it obj)

with the set of morphisms C(obj, obj) = M

the function * gives us the composition rule

and e in M gives us the identity morphism.

The Unit and associative laws of the Moniod correspond to the unit and associative laws of the category.

`a: Unit holds because \\(0 + m = m \\) and \\(m + 0 = m \\), associativity holds because \\((m_1 + m_2) + m_3 = m_1 + (m_2 + m_3) \\) b: Unit holds because \\([] ++ m = m \\) and \\( m ++ [] = m \\) associativity holds because \\((m_1 ++ m_2) ++ m_3 = m_1 ++ (m_2 ++ m_3) \\) c: A monoid can be viewed as a category with a single object (call it obj) with the set of morphisms C(obj, obj) = M the function * gives us the composition rule and e in M gives us the identity morphism. The Unit and associative laws of the Moniod correspond to the unit and associative laws of the category.`