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

- All Categories 2.3K
- Chat 500
- Study Groups 19
- Petri Nets 9
- Epidemiology 4
- Leaf Modeling 1
- Review Sections 9
- MIT 2020: Programming with Categories 51
- MIT 2020: Lectures 20
- MIT 2020: Exercises 25
- MIT 2019: Applied Category Theory 339
- MIT 2019: Lectures 79
- MIT 2019: Exercises 149
- MIT 2019: Chat 50
- UCR ACT Seminar 4
- General 68
- Azimuth Code Project 110
- Statistical methods 4
- Drafts 2
- Math Syntax Demos 15
- Wiki - Latest Changes 3
- Strategy 113
- Azimuth Project 1.1K
- - Spam 1
- News and Information 147
- Azimuth Blog 149
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 713

Options

*Is it an isomorphism?*

Suppose that someone tells you that their category \(C \) has two objects \(c,d \) and two non-identity morphisms, \(f : c \to d \) and \(g : d \to c \), but no other morphisms. Does \(f \) have to be the inverse of \(g \), i.e. is it forced by the category axioms that \( g◦f = id_c \) and \(f◦g= id_d \)?

## Comments

My argument is as follows :

(i) \(f◦g\) is a morphism from \(d \to d \). Since the only morphism in \(C \) from \(d \to d \) is \(id_d \) we must have \(f◦g= id_d \).

(ii) \(g◦f\) is a morphism from \(c \to c \). Since the only morphism in \(C \) from \(c \to c \) is \(id_c \) we must have \(g◦f= id_c \).

`My argument is as follows : (i) \\(f◦g\\) is a morphism from \\(d \to d \\). Since the only morphism in \\(C \\) from \\(d \to d \\) is \\(id_d \\) we must have \\(f◦g= id_d \\). (ii) \\(g◦f\\) is a morphism from \\(c \to c \\). Since the only morphism in \\(C \\) from \\(c \to c \\) is \\(id_c \\) we must have \\(g◦f= id_c \\).`