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.
Comment Source: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.
Comment Source: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.
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.