Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...

More documents

Recommendations

Info

150Now we compute(hQ) ◦ ( Qχ ) ◦ ( δ D QQ ) ? = ( P µ B Q)◦ (jB) ◦(DσB ) ◦ (DhQ)(hQ) ◦ ( Qχ ) ◦ ( δ D QQ )= (B µ P Q ) ◦ ( σ B P Q ) ◦ (P iQ) ◦ (qQ) ◦ ( Qχ ) ◦ ( δ D QQ )(164)= (B µ P Q ) ◦ ( ) (BP µ B Q ◦ BP µBQ B ) ◦ ( BP Qσ B B )◦ ( BP QP Qσ B) ◦ (BP QP iQ) ◦ (BP QqQ) ◦ ( u B P QQQ ) ◦ ( jQQ )Qmodfunctor=( Bµ P Q ) ◦ ( BP µ B Q)◦ (BP QmB ) ◦ ( BP Qσ B B ) ◦ ( BP QP Qσ B)◦ (BP QP iQ) ◦ (BP QqQ) ◦ ( u B P QQQ ) ◦ ( jQQ )u= (B Bµ P Q ) ◦ (u B P Q) ◦ ( P µ Q) B ◦ (P QmB ) ◦ ( P Qσ B B ) ◦ ( P QP Qσ B)◦ (P QP iQ) ◦ (P QqQ) ◦ ( jQQ )Qmodfunctor ( )= P µBQ ◦ (P QmB ) ◦ ( P Qσ B B ) ◦ ( P QP Qσ B) ◦ (P QP iQ)◦ (P QqQ) ◦ ( jQQ )σ= ( ) BP µ B Q ◦ (P QmB ) ◦ ( P QBσ B) ◦ ( P Qσ B P Q ) ◦ (P QP iQ)◦ (P QqQ) ◦ ( jQQ )(81)= ( P µ B Q)◦(P QσB ) ◦ ( P Q B µ P Q ) ◦ ( P Qσ B P Q ) ◦ (P QP iQ)◦ (P QqQ) ◦ ( jQQ )j= ( P µ B Q)◦ (jB) ◦(DσB ) ◦ ( D B µ P Q ) ◦ ( Dσ B P Q ) ◦ (DP iQ) ◦ (DqQ)= ( P µ B Q)◦ (jB) ◦(DσB ) ◦ (DhQ)Then the diagram above serially commute and hence in particularso thatπ Z ◦ (hQ) ◦ ( Qχ ) ◦ ( δ D QQ ) = π Z ◦ ( P µ B Q)◦ (jB) ◦(DσB ) ◦ (DhQ)π Z= π Z ◦ ( ε D P Q ) ◦ (DhQ) = π Z ◦ (hQ) ◦ ( ε D QQ )π Z ◦ (hQ) ◦ ( Qχ ) ◦ ( δ D QQ ) = π Z ◦ (hQ) ◦ ( ε D QQ ) .Since (B ′ , y ′ ) = Coequ Fun((Qχ)◦(δD QQ ) , ε D QQ ) , by the universal property ofcoequalizers, there exists a unique functorial morphism ν Z : B ′ → Z such that(167) ν Z ◦ y ′ = π Z ◦ (hQ) .We want to prove that ν Z is an isomorphism. Since hQ is an isomorphism, thereexists (hQ) −1 : P Q → QQ. Note that fromwe deduce that(hQ) ◦ ( Qχ ) ◦ ( δ D QQ ) = ( P µ B Q)◦ (jB) ◦(DσB ) ◦ (DhQ)(hQ) −1 ◦ (hQ) ◦ ( Qχ ) ◦ ( δ D QQ ) ◦ ( D (hQ) −1) == (hQ) −1 ◦ ( ) (P µ ) B Q ◦ (jB) ◦ DσB◦ (DhQ) ◦ ( D (hQ) −1)
151that is(168)Similarly, fromwe deduce that( ) (Qχ ◦ δD QQ ) ◦ ( D (hQ) −1) = (hQ) −1 ◦ ( ) (P µ ) B Q ◦ (jB) ◦ DσB.(ε D P Q ) ◦ (DhQ) = (hQ) ◦ ( ε D QQ )(169) (hQ) −1 ◦ ( ε D P Q ) = ( ε D QQ ) ◦ ( D (hQ) −1) .Thus we haveso thaty ′ ◦ (hQ) −1 ◦ ( P µ B Q)◦ (jB) ◦(DσB ) (168)= y ′ ◦ ( Qχ ) ◦ ( δ D QQ ) ◦ ( D (hQ) −1)defy ′= y ′ ◦ ( ε D QQ ) ◦ ( D (hQ) −1) (169)= y ′ ◦ (hQ) −1 ◦ ( ε D P Q )y ′ ◦ (hQ) −1 ◦ ( P µ B Q)◦ (jB) ◦(DσB ) = y ′ ◦ (hQ) −1 ◦ ( ε D P Q ) .(( ) (Since (Z, π Z ) = Coequ ) Fun P µBQ ◦ (jB) ◦ DσB, ε D P Q ) , by the universal propertyof coequalizers, there exists a unique functorial morphism ν Z′ : Z → B′ suchthat(170) ν ′ Z ◦ π Z = y ′ ◦ (hQ) −1 .Now we want to prove that ν ′ Z is the two-sided inverse of ν Z. Let us computeand since y ′ is an epimorphism we getMoreoverν ′ Z ◦ ν Z ◦ y ′ (167)= ν ′ Z ◦ π Z ◦ (hQ)(170)= y ′ ◦ (hQ) −1 ◦ (hQ) = y ′ν ′ Z ◦ ν Z = Id B ′.ν Z ◦ ν ′ Z ◦ π Z(170)= ν Z ◦ y ′ ◦ (hQ) −1 (167)= π Z ◦ (hQ) ◦ (hQ) −1 = π Zand since π Z is an epimorphism we deduce thatν Z ◦ ν ′ Z = Id Z .Thus ν Z is a functorial isomorphism between B ′ and Z with inverse ν Z ′ . Now we wantto construct an isomorphism between B and Z. Consider the parallel pairDP Q(P µ S Q )◦(jS)◦(Dσ S) (ε D P Q) P Qand computeσ B ◦ ( ) (P µ ) B Q ◦ (jB) ◦ DσB (81),j= m B ◦ ( σ B B ) ◦ ( P Qσ B) ◦ (jP Q)σ= B m B ◦ ( Bσ B) ◦ ( σ B P Q ) ◦ (jP Q) (67)= m B ◦ ( Bσ B) ◦ (u B P Q) ◦ ( ε D P Q )u B= m B ◦ (u B B) ◦ σ B ◦ ( ε D P Q ) Bmonad= σ B ◦ ( ε D P Q ) .
Page 1 and 2:
Contents1.
Page 3 and 4:
linearity and compatibility conditi
Page 5 and 6:
5and since g ◦ f is an epimorphis
Page 7 and 8:
Proof. Clearly (qP )◦(αP ) = (qP
Page 9 and 10:
Lemma 2.13 ([BM, L
Page 11 and 12:
i.e. Hom B (Y, iX) equalizes Hom B
Page 13 and 14:
13such thatd 0 ◦ v = Id Yd 1 ◦
Page 15 and 16:
f ↦→ Rfis bijective for every X
Page 17 and 18:
Remark 3.10. Let A = (A, m A , u A
Page 19 and 20:
19and fromµ A P ◦ ( µ A P A ) =
Page 21 and 22:
and thusk ◦ (u A QZ) ◦ (Qz) = h
Page 23 and 24:
and since A preserves equalizers, A
Page 25 and 26:
Conversely, let Φ be a functorial
Page 27 and 28:
Proof. Apply Proposition 3.24 to th
Page 29 and 30:
Since Q is a left A-module functor,
Page 31 and 32:
(Q BB F, p QB F ) = Coequ Fun(µBQ
Page 33 and 34:
Theorem 3.37. Let B = (B, m B , u B
Page 35 and 36:
where A UG B F : B → A is such th
Page 37 and 38:
Proposition 3.44. Let A = (A, m A ,
Page 39 and 40:
Note that, since f and g are A-bili
Page 41 and 42:
Proposition 3.54. Let (L, R) be an
Page 43 and 44:
Corollary 3.58. Let (L, R) be an ad
Page 45 and 46:
Definition 4.2. A
Page 47 and 48:
Proposition 4.13. Let C = ( C, ∆
Page 49 and 50:
Then we have(P Cx) ◦ ( ρ C P X )
Page 51 and 52:
and since C preserves coequalizers,
Page 53 and 54:
Proof. Apply Corollary 4.24 to the
Page 55 and 56:
Let( (CQ ) ()D, ι Q) C = Equ Fun
Page 58 and 59:
58F D right D-comodule functors Q :
Page 60 and 61:
60prove that C ν D : C F D → (C
Page 62 and 63:
624.2. The compari
Page 64 and 65:
64and[(Ω ◦ Γ) (ϕ)] (Y ) = (LY,
Page 66 and 67:
66for every ( X, C ρ X)∈ C A, th
Page 68 and 69:
68i.e.(44) (d ϕ K ϕ Y ) ◦ (̂η
Page 70 and 71:
70In particular(49) d ϕ(CX, ∆ C
Page 72 and 73:
72We have to prove that (LD ϕ , Ld
Page 74 and 75:
74we have that Ld ϕ K ϕ Y is mono
Page 76 and 77:
and since d is mono we get that(ε
Page 78 and 79:
78Corollary 4.63 (Beck’s Precise
Page 80 and 81:
80We compute(LRɛLY ′ ) ◦ ( LR
Page 82 and 83:
82Proof. First of all we prove that
Page 84 and 85:
84i.e. Aα is a functorial morphism
Page 86 and 87:
86Then we haveA µ CCX ◦ ( A∆ C
Page 88 and 89:
884.23) is a functor Ã : C A → C
Page 90 and 91:
90Let θ l = ( σ B P Q ) ◦ (P τ
Page 92 and 93:
925)σ A = ( ε C A ) ◦ ( Cσ A)
Page 94 and 95:
94(ii) the functorial morphism can
Page 96 and 97:
96defΦ= ( QP A µ Q)◦(QP σ A Q
Page 98 and 99:
98AU A can AA F = can AA F = ( CσA
Page 100 and 101: 100Similarly, one can prove the sta
Page 102 and 103: 102(b) A comonad C = ( C, ∆ C ,
Page 104 and 105: 104We calculateso that we getx ◦
Page 106 and 107: 106There exist functorial morphisms
Page 108 and 109: 108andsatisfying(B, y) = Coequ Fun(
Page 110 and 111: 1104) With notations of Theorem 6.2
Page 112 and 113: 112Then ν : Y → D is the unique
Page 114 and 115: 114= A µ Q ◦ ( Aε C Q ) ◦ (AC
Page 116 and 117: 116= ( Aε C Q ) ◦ ( cocan1 −1
Page 118 and 119: 118so that we getχ= (Cx) ◦ (C ρ
Page 120 and 121: 120We want to prove that Γ is an o
Page 122 and 123: 122and since Dε D is an epimorphis
Page 124 and 125: 124χ= (Cχ) ◦ (C ρ Q P Q ) ◦
Page 126 and 127: 126Now, since cocan 1 : AC → QP i
Page 128 and 129: 1287. Herds and Coherds7.1.
Page 130 and 131: 130◦ ( σ A QQQ ) ◦ (A µ Q P Q
Page 132 and 133: 132= µ B Q ◦ (A µ Q B ) ◦ ( A
Page 134 and 135: 134Assume now that there is another
Page 136 and 137: 136and hence we get(160) x ◦ (χP
Page 138 and 139: 138Proposition 7.7. In the setting
Page 140 and 141: 140We calculateA µ Q ◦ ( σ A Q
Page 142 and 143: 142x=and=δ C=(l= QlQ ̂QQ)◦ (QP
Page 144 and 145: 144(◦ ρ D ̂QQ)Q◦ (QlQ) ◦ (Q
Page 146 and 147: 146given byWe computeσ B = m B ◦
Page 148 and 149: 148andy= ′m B ◦ (ν B B) ◦ (y
Page 152 and 153: 152Thus we obtainσ B ◦ ( ) (P µ
Page 154 and 155: 154Thus hQ is an isomorphism with i
Page 156 and 157: 156) ( )l=(pb Q AQ B ◦ ̂QA µ QB
Page 158 and 159: 158In fact we haveTherefore we dedu
Page 160 and 161: 160χ= h 1 ◦ (P xQ B ) ◦ (P QP
Page 162 and 163: 162so that we obtain:(190)We comput
Page 164 and 165: 164(194)=) )(p QB ̂QA ◦(Qpb Q◦
Page 166 and 167: 166= Ξ ◦ (A A U A λ) ◦ (xx A
Page 168 and 169: 168)(155)= k 2 ◦(Qpb Q◦ (Ql A U
Page 170 and 171: 170) ) (χ= ρ ◦(p QB ̂QA ◦(Qp
Page 172 and 173: 172Theorem 8.13. Let A and B be cat
Page 174 and 175: 174l = eC ρ L : L = − ⊗ B A
Page 176 and 177: and[µBQ ◦ ( Qσ B)] (− ⊗ T x
Page 178 and 179: 178so that− ⊗ R 1 A ⊗ R c = (
Page 180 and 181: 180− ⊗ T x ⊗ R 1 A ⊗ A f
Page 182 and 183: 182(208)(209)(210)(211)(h1 ) 0 ⊗
Page 184 and 185: 184= abd 0 ⊗ d 1 1 ⊗ d 2 1b⊗d
Page 186 and 187: 186so that h 1 ⊗ h 2 ⊗ a ∈ A
Page 188 and 189: 188= 〈( h (1) y (1))εH ( h (2) y
Page 190 and 191: 190H C is faithfully coflat. Assume
Page 192 and 193: 192=(Qε C H C) ( ∑ )−□ C k i
Page 194 and 195: 194Following Theorem 6.29, we now c
Page 196 and 197: 196)û E(ε C H C (h) = û E (π (h
Page 198 and 199: 198Letandα l = (ϕ ⊗ H) ( (x ⊗
Page 200 and 201:
200This map is well-defined, in fac
Page 202 and 203:
202We now have to prove that this m
Page 204 and 205:
2041) − ⊗ B Σ A preserves the
Page 206 and 207:
206functorial isomorphism. In parti
Page 208 and 209:
208coaction ρ C Σ : Σ → Σ ⊗
Page 210 and 211:
210Now, we consider a particular ca
Page 212 and 213:
212Definition 9.27. Let k be a comm
Page 214 and 215:
214∆coass= a ⊗ c (1) ⊗ A 1 A
Page 216 and 217:
216Definition 9.32.</strong
Page 218 and 219:
218Let us compute, for every d ∈
Page 220 and 221:
220• 2-cells: monad functor trans
Page 222 and 223:
222We now want to prove that ρ Q·
Page 224 and 225:
224Proof. Let us consider the follo
Page 226 and 227:
226and since p Q•B Q ′ ,Q ′
Page 228 and 229:
228(241)= (1 Q • B l Q ′) ζ Q,
Page 230 and 231:
230On the other hand, we can first
Page 232 and 233:
232so that we define the map φ F (
Page 234 and 235:
234Since we have(B • B (Q · A) ,
Page 236 and 237:
2362-cells. This means that a comon
Page 238 and 239:
238defined by settingu Q·A = ( u (
Page 240 and 241:
240the unique A-bimodule morphism s
Page 242 and 243:
242Let F be a finite subset of Hom
Page 244 and 245:
244Lemma A.4. Let A be an abelian c
Page 246 and 247:
246We haveT (ζ) ◦ ξ ◦ T H (p)
Page 248 and 249:
248where k : Ker (Coker (f ◦ p))
Page 250 and 251:
250be the codiagonal map of the ρ
Page 252 and 253:
252Proposition A.12 ([ELGO2, Propos
Page 254 and 255:
254(⇒) Let {A i } i∈Ibe a famil
Page 256 and 257:
256We will prove that h : ∐ B i
Page 258 and 259:
258Proposition A.19. Let (T, H) be
Page 260 and 261:
260Since P is finite Hom A (P, P )
Page 262 and 263:
262andP (J ′ )e f ′−→ P (I
Page 264 and 265:
264hence there exists a unique morp
Page 266:
266[RW] R. Rosebrugh, R.J. Wood, Di
show all

Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...

Create successful ePaper yourself

Delete template?

Save as template?