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Preorders.
A preorder is a category such that, for every two objects \(a,b \), there is at most one morphism \(a \to b \). That is, there either is or is not a morphism from \(a \) to \(b \), but there are never two morphisms \(a \) to \(b \). If there is a morphism \(a \to b \), we write \(a \leq b \); if there is not a morphism \(a \to b \), we don’t. For example, there is a preorder \(P \) whose objects are the positive integers \(Ob(P) = \mathbb{N}_{≥1} \) and whose hom-sets are given by \(P(a,b) := \{x \in N | x∗a = b \} \) This is a preorder because either \(P(a,b) \) is empty (if \(b \) is not divisible by \(a \)) or contains exactly one element.
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(a) What is the identity on 12?
(b) Show that if \(x : a \to b \) and \(y : b \to c \) are morphisms, then there is a morphism \(y◦x \) to serve as their composite.
(c) Would it have worked just as well to take \(P \) to have all of \(\mathbb{N} \) as objects, rather than just the positive integers?
Comments
a) 1 (1*12 = 12)
b) xa = b and yb = c, thus yxa = c
c) no, we would have more than one morphism from
0 -> 0a) 1 (1*12 = 12) b) x*a = b and y*b = c, thus y*x*a = c c) no, we would have more than one morphism from `0 -> 0`If a =2 and b = 3 what is x?
If a =2 and b = 3 what is x?@JimStuttard there is no x, so there is no morphism from 2 to 3
@JimStuttard there is no x, so there is no morphism from 2 to 3@FabricioOlivetti Duh. Tnx. My bad, somehow missed the divisibility part of the definition.
@FabricioOlivetti Duh. Tnx. My bad, somehow missed the divisibility part of the definition.