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Table of Contents

  1. 67.5.7
  2. Problem 67A
  3. Problem 67B
  4. Problem 67C

67.5.7 #

Exercise 67.5.7. Show that in 𝐒𝐞𝐭, 𝐆𝐫𝐩, 𝐀𝐛, 𝐕𝐞𝐜𝐭𝑘, 𝐓𝐨𝐩, the notions of epic and surjective coincide. (For𝐒𝐞𝐭, take 𝐴={0,1}.)Solution by finalchildConsider an arrow 𝑓:𝑋→𝑌 in the category. The proof ⇐ direction is almost the same for all concretecategories, so it’s only written once.𝐒𝐞𝐭⇒) ∀𝑦∈𝑌, consider 𝑔:𝑌→{0,1}=𝑎↦0 and ℎ:𝑌→{0,1}=𝑎↦(𝑎=?𝑦). Since 𝑓 is epic, 𝑔∘𝑓≠ℎ∘𝑓. Thus, ∃𝑥∈𝑋. 𝑓(𝑥)=𝑦. ∎⇐) ∀𝐴∈𝐒𝐞𝐭,∀𝑔:𝑌→𝐴,∀𝑦∈𝑌. Since 𝑓 is surjective, ∃𝑥∈𝑋.𝑓(𝑥)=𝑦. 𝑔(𝑦)=(𝑔∘𝑓)(𝑥). Thus, 𝑔 isdetermined by 𝑔∘𝑓. ∎𝐆𝐫𝐩⇒) Cannot find proof with learnt concepts.𝐀𝐛⇒) Consider 𝑔:𝑌→𝑌/Im𝑓=𝑎↦1Im𝑓 and ℎ:𝑌→𝑌/Im𝑓=𝑎↦𝑎Im𝑓. Since 𝑓 is epic and 𝑔∘𝑓=ℎ∘𝑓, 𝑔=ℎ. Thus, Im𝑓=𝑌. ∎𝐕𝐞𝐜𝐭𝑘⇒) Consider 𝑔:𝑌→𝑌/Im𝑓=𝑎↦0+Im𝑓 and ℎ:𝑌→𝑌/Im𝑓=𝑎↦𝑎+Im𝑓. Since 𝑓 is epic and𝑔∘𝑓=ℎ∘𝑓, 𝑔=ℎ. Thus, Im𝑓=𝑌. ∎𝐓𝐨𝐩⇒) Consider 𝑔:𝑌→𝑌/Im𝑓=𝑎↦[Im𝑓] and ℎ:𝑌→𝑌/Im𝑓=𝑎↦[𝑎]. Since 𝑓 is epic and 𝑔∘𝑓=ℎ∘𝑓, 𝑔=ℎ. Thus, Im𝑓=𝑌. ∎

Problem 67A #

Problem 67A. In the category 𝖵𝖾𝖼𝗍𝑘 of 𝑘-vector spaces (for a field 𝑘), what are the initial and terminalobjects?Solution by RanolPThe arrow of 𝖵𝖾𝖼𝗍𝑘 is a Linear Map, which preserves Vector Addition and Scalar Multiplication.Claim.The initial is {0}Proof.Suppose there’s an arrow 𝑓:{0}→𝑊, since it’s linear map, it should respect𝑓(𝑐𝐯)=𝑐𝑓(𝐯)Thus, for 𝑐=0 and 𝑓(𝐯)=𝐰, 𝑓(0𝑉)=𝑓(0𝐯)=0𝑓(v)=0𝐰=0𝑊.Because the domain is {0}, there is no way to make a linear map other than 𝑓(0𝑉)=0𝑊.Hence, there’s only one arrow from {0} to the other spaces. ∎Claim.The terminal is {0}Proof.Suppose there’s an arrow 𝑓:𝑉→{0}. Since the codomain is {0}, The function must be 𝑓(𝐯)=0,which is trivially linear map.Hence, there’s only one arrow to {0} from the other spaces. ∎

Problem 67B #

Problem 67B†. What is the coproduct 𝑋+𝑌 in the categories 𝖲𝖾𝗍, 𝖵𝖾𝖼𝗍𝑘, and a poset?Solution by kiwiyouCoproduct in 𝖲𝖾𝗍Define the coproduct as𝑋+𝑌=𝑋Π𝑌𝜄𝑋(𝑥)=(𝑥,0)𝜄𝑌(𝑦)=(𝑦,1)Then for any object 𝐴 with morphisms 𝑔:𝑋→𝐴 and ℎ:𝑌→𝐴, 𝑓:𝑋+𝑌→𝐴 must satisfy(𝑓∘𝜄𝑋)(𝑥)=𝑓((𝑥,0))=𝑔(𝑥)(𝑓∘𝜄𝑌)(𝑦)=𝑓((𝑦,1))=ℎ(𝑦)This uniquely determines 𝑓 for all (𝑥,𝑦)∈𝑋+𝑌.∎Coproduct in 𝖵𝖾𝖼𝗍𝑘Define the coproduct as𝑋+𝑌={(𝐱,𝐲)|𝐱∈𝑋,𝐲∈𝑌}𝑘(𝐱,𝐲)=(𝑘𝐱,𝑘𝐲)(𝐱1,𝐲1)+(𝐱2,𝐲2)=(𝐱1+𝐱2,𝐲1+𝐲2)𝜄𝑋(𝐱)=(𝐱,𝟎𝑌)𝜄𝑌(𝐲)=(𝟎𝑋,𝐲)Then for any object 𝐴 with morphisms 𝑔:𝑋→𝐴 and ℎ:𝑌→𝐴, 𝑓:𝑋+𝑌→𝐴 must satisfy(𝑓∘𝜄𝑋)(𝐱)=𝑓((𝐱,𝟎𝑌))=𝑔(𝐱)(𝑓∘𝜄𝑌)(𝐲)=𝑓((𝟎𝑋,𝐲))=ℎ(𝐲)Since 𝑓((𝐱,𝐲))=𝑓((𝐱,𝟎𝑌))+𝑓((𝟎𝑋,𝐲))=𝑔(𝐱)+ℎ(𝐲), this uniquely determines 𝑓 for all (𝐱,𝐲)∈𝑋+𝑌.∎Coproduct in a posetDefine the coproduct as𝑋+𝑌=𝑋∨𝑌if it exists.Then for any object 𝐴, 𝑋≤𝐴∧𝑌≤𝐴⟹𝑋+𝑌≤𝐴.∎

Problem 67C #

Problem 67C. In any category 𝒜 where all products exist, show that(𝑋×𝑌)×𝑍≅𝑋×(𝑌×𝑍)where 𝑋,𝑌,𝑍 are arbitrary objects. (Here both sides refer to the objects, as in Abuse of Notation 67.4.2.)Solution by Jihyeon Kim (김지현) (simnalamburt)목표: 아래를 만족하는 morphism 𝑓,𝑔를 찾는것이다.𝑓:(𝑋×𝑌)×𝑍→𝑋×(𝑌×𝑍)𝑔:𝑋×(𝑌×𝑍)→(𝑋×𝑌)×𝑍𝑔∘𝑓=id(𝑋×𝑌)×𝑍𝑓∘𝑔=id𝑋×(𝑌×𝑍)곱의 보편성에 의해, 임의의 C와 𝑔:𝐶→𝐴,ℎ:𝐶→𝐵가 주어지면𝑓:𝐶→𝐴×𝐵라는 유일한 화살표가 존재하여𝜋𝐴∘𝑓=𝑔𝜋𝐵∘𝑓=ℎ를 만족하는데, 편의상 𝑓를 ⟨𝑔,ℎ⟩로 표기하겠다.그리고 이제 아래의 아름다운 그림을 보라:𝑝≔𝜋𝑋×𝑌𝑋×𝑌,𝑍𝑞≔𝜋𝑍𝑋×𝑌,𝑍𝑟≔𝜋𝑌𝑌,𝑍𝑠≔𝜋𝑍𝑌,𝑍𝑢≔𝜋𝑋𝑋,𝑌×𝑍𝑣≔𝜋𝑌×𝑍𝑋,𝑌×𝑍𝛼≔𝜋𝑋𝑋,𝑌𝛽≔𝜋𝑌𝑋,𝑌⟨𝛽∘𝑝,𝑞⟩𝑓≔⟨𝛼∘𝑝,⟨𝛽∘𝑝,𝑞⟩⟩⟨𝑢,𝑟∘𝑣⟩𝑔≔⟨⟨𝑢,𝑟∘𝑣⟩,𝑠∘𝑣⟩(𝑋×𝑌)×𝑍𝑌×𝑍𝑋𝑌𝑍𝑋×𝑌𝑋×(𝑌×𝑍)