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

- All Categories 2.3K
- Chat 499
- Study Groups 18
- Petri Nets 9
- Epidemiology 3
- 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 67
- Azimuth Code Project 110
- Statistical methods 3
- 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 708

Options

## Comments

\(G_1\): \(f = g\)

\(G_2\): None (I think)

\(G_3\): \(f.h = g.i\)

\(G_4\): None

`\\(G_1\\): \\(f = g\\) \\(G_2\\): None (I think) \\(G_3\\): \\(f.h = g.i\\) \\(G_4\\): None`

For \(\mathcal{G}_2\), shouldn't we impose \(f\) to be the identity morphism, that is \(f = 1\)? Otherwise, there are infinitely many morphisms from \(\bullet\) to \(\bullet\): \(1, f, f \circ f, f \circ f \circ f\), etc.

`For \\(\mathcal{G}_2\\), shouldn't we impose \\(f\\) to be the identity morphism, that is \\(f = 1\\)? Otherwise, there are infinitely many morphisms from \\(\bullet\\) to \\(\bullet\\): \\(1, f, f \circ f, f \circ f \circ f\\), etc.`