Howdy, Stranger!
It looks like you're new here. If you want to get involved, click one of these buttons!
Categories
- All Categories 2.2K
- Applied Category Theory Course 355
- Applied Category Theory Seminar 4
- Exercises 149
- Discussion Groups 49
- How to Use MathJax 15
- Chat 480
- Azimuth Code Project 108
- News and Information 145
- Azimuth Blog 149
- Azimuth Forum 29
- Azimuth Project 189
- - Strategy 108
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 711
- - Latest Changes 701
- - - 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 21
- - - Statistical methods 2
- - - Sustainability 4
- - - Things to do 2
- - - Visualisation 1
- General 39
Comments
1) The monogamous spousal function for all people, where unmarried people are treated as their own spouse.
The spouse, \(s\), of my spouse, \(z_1\) is me, \(z_0\).
2) The a mother, \(a\), of only children, \(c\), where \(g\) is the firstborn and \(h\) the last.
1) The monogamous spousal function for all people, where unmarried people are treated as their own spouse. The spouse, \\(s\\), of my spouse, \\(z_1\\) is me, \\(z_0\\). 2) The a mother, \\(a\\), of only children, \\(c\\), where \\(g\\) is the firstborn and \\(h\\) the last.About 2 - it looks like the morphisms g and h should really be the same morphism, is there a way to show this? If we had a reverse morphism \(f^{-1}: b \to a\), such that \(f \cdot f^{-1} = id_b\) then this would follow naturally, but we don't. We just have some image of a, \(f: a \to Im(a)\), in b. So basically we are saying that this equality holds only for a subset of b, other entries may have different outcomes from g and h.
EDIT: got it, if a contains mothers with a single child, and b is the set of all mothers, then \(f.g = f.h\), but in general, for other mothers, the first child and the last are not equal.
About 2 - it looks like the morphisms **g** and **h** should really be the same morphism, is there a way to show this? If we had a reverse morphism \\(f^{-1}: b \to a\\), such that \\(f \cdot f^{-1} = id_b\\) then this would follow naturally, but we don't. We just have some image of *a*, \\(f: a \to Im(a)\\), in *b*. So basically we are saying that this equality holds only for a subset of *b*, other entries may have different outcomes from **g** and **h**. **EDIT**: got it, if _a_ contains mothers with a single child, and _b_ is the set of all mothers, then \\(f.g = f.h\\), but in general, for other mothers, the first child and the last are not equal.