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

- All Categories 2.2K
- Applied Category Theory Course 323
- Applied Category Theory Exercises 149
- Applied Category Theory Discussion Groups 43
- Applied Category Theory Formula Examples 15
- Chat 468
- Azimuth Code Project 107
- News and Information 145
- Azimuth Blog 148
- Azimuth Forum 29
- Azimuth Project 190
- - Strategy 109
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 707
- - Latest Changes 699
- - - Action 14
- - - Biodiversity 8
- - - Books 2
- - - Carbon 9
- - - Computational methods 38
- - - Climate 53
- - - Earth science 23
- - - Ecology 43
- - - Energy 29
- - - Experiments 30
- - - Geoengineering 0
- - - Mathematical methods 69
- - - Meta 9
- - - Methodology 16
- - - Natural resources 7
- - - Oceans 4
- - - Organizations 34
- - - People 6
- - - Publishing 4
- - - Reports 3
- - - Software 20
- - - Statistical methods 2
- - - Sustainability 4
- - - Things to do 2
- - - Visualisation 1
- General 38

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.`