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# Exercise 47 - Chapter 3

edited June 2018

Give an example of the sort of data that would make sense for the following schemas:

• Options
1.
edited May 2018

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. 
• Options
2.
edited June 2018

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.