Problem 68A#Problem 68A. Show that the two definitions of natural transformation (one in terms of 𝒜︀×𝟐→ℬ︀ and one in terms of arrows 𝐹(𝐴)⟶𝛼𝐴𝐺(𝐴)) are equivalent.Solution by RanolPCommon ConditionGiven two categories 𝒜︀,ℬ︀ and two functors 𝐹,𝐺:𝒜︀→ℬ︀.Given natural transformation 𝐹(𝐴)⟶𝛼𝐴𝐺(𝐴)Our goal is to construct a functor 𝛽:𝒜︀×𝟐→ℬ︀ for each 𝛼.Since product category is defined as tuple of objects and arrows, we can construct 𝛽 bysending tuple of objects𝛽((𝐴,0))=𝐹(𝐴)𝛽((𝐴,1))=𝐺(𝐴)and sending tuple of morphisms𝛽((𝑓,id0))=𝐹(𝑓)𝛽((𝑓,id1))=𝐺(𝑓)𝛽((𝐴1→𝑓𝐴2,0→≤1))=𝛼𝐴2∘𝐹(𝑓)toℬ︀Now we need to check the 𝛽 is functor.1.Mapping of ObjectsTrivially done by 𝐹(𝐴) and 𝐺(𝐴)2.Mapping of MorphismsTrivially done by 𝐹(𝑓), 𝐺(𝑓) and 𝛼𝐴2∘𝐹(𝑓).𝐹(𝑓)𝛼𝐴1𝛼𝐴2𝐺(𝑓)𝐹(𝐴1)𝐹(𝐴2)𝐺(𝐴1)𝐺(𝐴2)3.Preservation of Identity MorphismIn 𝒜︀×𝟐, identity is either id(𝐴,0)=(id𝐴,id0) which will be sent to 𝐹(id𝐴) or id(𝐴,1)=(id𝐴,id1) which will be sent to 𝐺(id𝐴). both will preserve identity by functor 𝐹 and 𝐺.4.Preservation of CompositionIt’s the most unsatisfying part. We will split cases for easier understanding.i.Both sides are (𝑓,id0) form: Preserved by functor 𝐹.ii.Both sides are (𝑓,id1) form: Preserved by functor 𝐺.iii.The one is (𝐴1→𝑓𝐴2,0→≤1): Prove case by case.a.𝛽((𝐴2→𝑔𝐴3,id1))∘𝛽((𝐴1→𝑓𝐴2,0→≤1))=𝐺(𝑔)∘𝛼𝐴2∘𝐹(𝑓)Abuse of Notation: let’s call the function above as 𝛽(𝑔)∘𝛽(𝑓) for now.𝑓𝑔∘𝑓𝑔𝐹(𝑓)𝛽(𝑔)∘𝛽(𝑓)𝛼𝐴2𝐺(𝑔)𝐴1𝐴2𝐴3𝐹(𝐴1)𝐹(𝐴2)𝐺(𝐴2)𝐺(𝐴3)The diagram above commutes.b.(𝐴1→𝑓𝐴2,0→≤1)∘(𝐴0→𝑔𝐴1,id0)=𝛼𝐴2∘𝐹(𝑓)∘𝐹(𝑔)Abuse of Notation: let’s call the function above as 𝛽(𝑓)∘𝛽(𝑔) for now.𝑔𝑓∘𝑔𝑓𝐹(𝑔)𝛽(𝑓)∘𝛽(𝑔)𝐹(𝑓)𝛼𝐴2𝐴0𝐴1𝐴2𝐹(𝐴0)𝐹(𝐴1)𝐹(𝐴2)𝐺(𝐴2)The diagram above commutes.Given functor 𝛽:𝒜︀×𝟐→ℬ︀Our goal is to construct a natural transformation 𝐹(𝐴)⟶𝛼𝐴𝐺(𝐴) from 𝛽.Consider 𝐴1→𝑓𝐴2, then following diagram commutes.(𝑓,id0)(id𝐴1,0→1)(id𝐴2,0→1)(𝑓,id1)𝛽𝐹(𝑓)𝛽((id𝐴1,0→1))𝛽((id𝐴2,0→1))𝐺(𝑓)(𝐴1,0)(𝐴2,0)(𝐴1,1)(𝐴2,1)𝐹(𝐴1)𝐹(𝐴2)𝐺(𝐴1)𝐺(𝐴2)Let’s say 𝛼𝐴=𝛽((id𝐴,0→1)). Then it’s the naturally requirement.𝐹(𝑓)𝛼𝐴1𝛼𝐴2𝐺(𝑓)𝐹(𝐴1)𝐹(𝐴2)𝐺(𝐴1)𝐺(𝐴2)
Problem 68B#Problem 68B. Let 𝒜︀ be the category of finite sets whose arrows are bijections between sets. For 𝐴∈𝒜︀, let 𝐹(𝐴) be the set of permutations of 𝐴 and let 𝐺(𝐴) be the set of orderings on 𝐴.1(a)Extend 𝐹 and 𝐺 to functors 𝒜︀→Set.(b)Show that 𝐹(𝐴)≅𝐺(𝐴) for every 𝐴, but this isomorphism is not natural.Solution by finalchild(a)𝐹 and 𝐺 act on arrows as follows:𝐹(𝑓)=𝑝↦𝑓∘𝑝∘𝑓−1𝐺(𝑓)=𝑜↦𝑓∘𝑜(b)|𝐹(𝐴)|=|𝐺(𝐴)|, hence 𝐹(𝐴)≅𝐺(𝐴).Consider 𝐴={0,1}, and an arrow 𝑓:𝐴→𝐴=𝑥↦1−𝑥. Let 𝛼𝐴 be an isomorphism from 𝐹(𝐴) to 𝐺(𝐴).Since conjugation has no effect on permutations on two-element set, 𝛼𝐴∘𝐹(𝑓)=𝛼𝐴. On the other hand, 𝐺(𝑓)∘𝛼𝐴=reverse∘𝛼𝐴. The two cannot be equal, thus this isomorphism is not natural.1A permutation is a bijection 𝐴→𝐴, and an ordering is a bijection {1,…,𝑛}→𝐴, where 𝑛 is the size of 𝐴.