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  1. 5.2.25
  2. 5.3.11
  3. 5.3.8

5.2.25 #

[Le14] 5.2.25. A map 𝑚:𝐴→𝐵 is regular monic if there exist an object 𝐶 and maps 𝐵⇉𝐶 of which 𝑚is an equalizer. A map 𝑚:𝐴→𝐵 is split monic if there exists a map 𝑒:𝐵→𝐴 such that 𝑒𝑚=1𝐴.(a)Show that split monic ⟹ regular monic ⟹ monic.(b)In 𝖠𝖻, show that all monics are regular but not all monics are split.(c)In 𝖳𝗈𝗉, describe the regular monics, and find a monic that is not regular.Solution by kiwiyou(a) Split Monic ⟹ Regular monicLet map 𝑚:𝐴→𝐵 be a split monic, i.e. 𝑒∘𝑚=1𝐴 for some 𝑒:𝐵→𝐴.We will show that 𝑚 is an equalizer of 𝑚∘𝑒:𝐵→𝐵 and 1𝐵:𝐵→𝐵.For some object 𝑂, suppose a map 𝑓:𝑂→𝐵 such that (𝑚∘𝑒)∘𝑓=1𝐵∘𝑓.Let 𝑢=𝑒∘𝑓. We need to show this 𝑢 is the only map that satisfies 𝑚∘𝑢=𝑓.If there exists a map 𝑣:𝑂→𝐴 such that 𝑚∘𝑣=𝑓, then𝑢=𝑒∘𝑓=𝑒∘(𝑚∘𝑣)=(𝑒∘𝑚)∘𝑣=1𝐴∘𝑣=𝑣.Therefore, 𝑢 is the only map that satisfies 𝑚∘𝑢=𝑓.∎(a) Regular Monic ⟹ MonicLet 𝑚:𝐴→𝐵 be a regular monic, i.e. 𝑚 is an equalizer for some 𝐶, 𝑓:𝐵→𝐶 and 𝑔:𝐵→𝐶.To show that 𝑚 is monic, suppose maps ℎ1:𝑂→𝐴 and ℎ2:𝑂→𝐴 both satisfies 𝑚∘ℎ1=𝑚∘ℎ2 for someobject 𝑂. We need to show that ℎ1=ℎ2.Note that 𝑚∘ℎ1=𝑚∘ℎ2 is a map from 𝑂 to 𝐶.Since 𝑚 is an equalizer, there exists a unique map ℎ:𝑂→𝐴 such that 𝑓∘𝑚∘ℎ=𝑔∘𝑚∘ℎ.Therefore ℎ1=ℎ2=ℎ.∎(b) In 𝖠𝖻, show that all monics are regular.In 𝖠𝖻, monic = injective homomorphism.Let 𝑚:𝐴→𝐵 be a monic. Since 𝑚img(𝐴) is a normal subgroup of 𝐵, we have 𝐶=𝐵/𝑚img(𝐴).Let 𝑓:𝐵→𝐶 be the quotient, and 𝑔:𝐵→𝐶 be the trivial homomorphism.Then for all ℎ:𝐷→𝐵 such that 𝑓∘ℎ=𝑔∘ℎ=1𝐶, 𝑢=𝑚−1∘ℎ is the unique homomorphism that satisfies𝑚∘𝑢=ℎ.Since 𝑚 is an equalizer, 𝑚 is regular monic.∎(b) In 𝖠𝖻, show that not all monics are split.Define 𝑚:ℤ→ℤ as 𝑚(𝑛)=2𝑛. This homomorphism is injective, thus monic.Suppose 𝑚 is split monic, i.e. 𝑒∘𝑚=1 for some homomorphism 𝑒:ℤ→ℤ. This contradicts because:1=(𝑒∘𝑚)(1)=𝑒(2)=𝑒(1)+𝑒(1).Therefore, 𝑚 is not split monic.∎

5.3.11 #

Le14 5.3.11. It was shown in Example 5.3.4 that the forgetful functor 𝑈:𝐆𝐫𝐩→𝐒𝐞𝐭 creates binaryproducts.(a)Using the formula for limits in 𝐒𝐞𝐭 (Example 5.1.22), prove that, in fact, 𝑈 creates arbitrary limits.(b)Satisfy yourself that the same is true if 𝐆𝐫𝐩 is replaced by any other category of algebras such as 𝐑𝐢𝐧𝐠,𝐀𝐛, or 𝐕𝐞𝐜𝐭𝑘Solution by finalchild(a)∀𝐽∈𝐂𝐚𝐭 ∀𝐹:𝐽→𝐆𝐫𝐩.A unique cone (𝐿′,𝜑′) to 𝐹 where 𝑈(𝐿′)=𝐿 and 𝑈(𝜑′𝑋)=𝜑𝑋 exists.Let a cone (𝐿,𝜑) be the limit of 𝑈∘𝐹, constructed using Example 5.1.22. This exists since all limitsexist in 𝐒𝐞𝐭.Let a cone (𝐿′,𝜑′) to 𝐹 where 𝑈(𝐿′)=𝐿 and 𝑈(𝜑′𝑋)=𝜑𝑋. This says 𝐿′ is just 𝐿 with a group operation⋅𝐿′ and 𝜑′𝑋=𝜑𝑋.Since 𝜑𝑋 is a group homomorphism, 𝜑𝑋(𝑙1⋅𝐿′𝑙2)=𝜑𝑋(𝑙1)⋅𝐹(𝑋)𝜑𝑋(𝑙2)Thus, 𝑙1⋅𝐿′𝑙2=(𝜑𝑋(𝑙1)⋅𝐹(𝑋)𝜑𝑋(𝑙2))𝑋∈𝐽, which is closed in 𝐿 since for all 𝑓:𝑋→𝑌 in𝐽, 𝑈(𝑓)(𝜑𝑋(𝑙1⋅𝐿′𝑙2))=𝑈(𝑓)(𝜑𝑋(𝑙1)⋅𝐹(𝑋)𝜑𝑋(𝑙2))=𝑈(𝑓)(𝜑𝑋(𝑙1))⋅𝐹(𝑌)𝑈(𝑓)(𝜑𝑋(𝑙2))=𝜑𝑌(𝑙1)⋅𝐹(𝑌)𝜑𝑌(𝑙2)=𝜑𝑌(𝑙1⋅𝐿′𝑙2) satisfying the condition for inclusion in 𝐿.)Similarly, it can be shown that the inverse is closed in 𝐿′.This makes the unique group 𝐿′, and it can be verified that 𝐿′ is a group and each of 𝜑 is a morphismin 𝐆𝐫𝐩 indeed. Moreover, (𝐿′,𝜑) is a cone in 𝐆𝐫𝐩 (trivial since (𝐿,𝜑) is a cone in 𝐒𝐞𝐭).(𝐿′,𝜑′) is a limit cone.Let a cone (𝑁,𝜓) to 𝐹. Let 𝑢:𝑁→𝐿′ such that ∀𝑋∈𝐽 𝜑′𝑋∘𝑢=𝜓𝑋. It immediately follows that𝑢(𝑛)=(𝜓𝑋(𝑛))𝑋∈𝐽It can be verified that 𝑢 is indeed a group homomorphism.𝑢(𝑛1⋅𝑁𝑛2)=(𝜓𝑋(𝑛1⋅𝑁𝑛2))𝑋∈𝐽=(𝜓𝑋(𝑛1)⋅𝐹(𝑋)𝜓𝑋(𝑛2))𝑋∈𝐽=(𝜑′𝑋(𝑢(𝑛1))⋅𝐹(𝑋)𝜑′𝑋(𝑢(𝑛2)))𝑋∈𝐽=𝑢(𝑛1)⋅𝐿′𝑢(𝑛2).Thus, 𝑢 is the unique arrow in 𝐆𝐫𝐩 factoring (𝑁,𝜓)(b)Satisfied myself.

5.3.8 #

Le14 5.3.8.Let 𝒜 be a category with binary products. Suppose that we have chosen for each pair (𝑋,𝑌) of objects aproduct cone𝑋⟵𝑝𝑋,𝑌1𝑋×𝑌⟶𝑝𝑋,𝑌2𝑌.Construct a functor 𝒜×𝒜→𝒜 given on objects by (𝑋,𝑌)↦𝑋×𝑌.Solution by RanolPLet 𝐹:𝒜×𝒜→𝒜 be a functor.Given condition, we send objects by 𝐹((𝑋,𝑌))=𝑋×𝑌.And we send morphisms (𝑋,𝑌)⟶(𝑓,𝑔)(𝑍,𝑊) by𝑓∘𝑝𝑋,𝑌1𝑔∘𝑝𝑋,𝑌2𝑝𝑋,𝑌1∃!𝐹((𝑓,𝑔))𝑓𝑝𝑋,𝑌2𝑔𝑝𝑍,𝑊1𝑝𝑍,𝑊2𝑋×𝑌𝑋𝑌𝑍×𝑊𝑍𝑊Now all we need to do is to check that 𝐹 is functor.1.Preservation of IdentityLet’s think about (𝑋,𝑌)⟶id(𝑋,𝑌)(𝑋,𝑌). Substitute the diagram above we getid𝑋∘𝑝𝑋,𝑌1id𝑌∘𝑝𝑋,𝑌2𝑝𝑋,𝑌1∃!𝐹((id𝑋,id𝑌))id𝑋𝑝𝑋,𝑌2id𝑌𝑝𝑋,𝑌1𝑝𝑋,𝑌2𝑋×𝑌𝑋𝑌𝑋×𝑌𝑋𝑌Let’s place id𝑋×𝑌 on the position of 𝐹((id𝑋,id𝑌)), then we get the desired properties. And since themorphism is unique, there’s no way to construct morphism other than id𝑋×𝑌. Thus the functor 𝐹 preservesidentity.2.Preservation of CompositionLet’s say we have two morphisms (𝑋,𝑌)⟶(𝑓1,𝑔1)(𝑍,𝑊) and (𝑍,𝑊)⟶(𝑓2,𝑔2)(𝐴,𝐵).𝑓2∘𝑓1∘𝑝𝑋,𝑌1𝑔2∘𝑔1∘𝑝𝑋,𝑌2𝑝𝑋,𝑌1𝐹((𝑓1,𝑔1))𝐹((𝑓2∘𝑓1,𝑔2∘𝑔1))𝑓1𝑝𝑋,𝑌2𝑔1𝐹((𝑓2,𝑔2))𝑝𝑍,𝑊1𝑓2𝑝𝑍,𝑊2𝑔2𝑝𝐴,𝐵1𝑝𝐴,𝐵2𝑋×𝑌𝑋𝑌𝑍×𝑊𝑍𝑊𝐴×𝐵𝐴𝐵Since the morphism 𝐹((𝑓2∘𝑓1,𝑔2∘𝑔1)) satisfying desired properties is unique so𝐹((𝑓2∘𝑓1,𝑔2∘𝑔1))=𝐹((𝑓2,𝑔2))∘𝐹((𝑓1,𝑔1))