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

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36satisfyingf B ◦ p P = p Q ◦ (f B U) .Then we can considerAf B : A P B → A Q Bsuch thatAU A f B = f B .Proof. Consider the following diagramP B B Uµ B P BUP B Uλ B P B U p PP Bf BfB B UQB B Uµ B QBUf B UQ B Uλ B Q B U pQ Q BSince f is a functorial morphism and it is a functorial morphism of right B-modulefunctors, the left square serially commutes. Note thatp Q ◦ (f B U) ◦ ( µ B P BU ) = p Q ◦ (f B U) ◦ (P B Uλ B )so that, by the universal property of the coequalizer, there exists a unique morphismf B : P B → Q B such that(18) f B ◦ p P = p Q ◦ (f B U) .We now want to prove that f B is a functorial morphism of left A-module functor.In fact we havef B ◦ A µ PB ◦ (Ap P ) (7)= f B ◦ p P ◦ (A µ P B U )(18)= p Q ◦ (f B U) ◦ (A µ P B U ) fleftAlin= p Q ◦ (A µ QB U ) ◦ (Af B U)(7)= A µ QB ◦ (Ap Q ) ◦ (Af B U) (18)= A µ QB ◦ (Af B ) ◦ (Ap P )and since A preserves coequalizers Ap Q is an epimorphism so that we getf B ◦ A µ PB = A µ QB ◦ (Af B ) .Then there exists a functorial morphism A f B : A P B → A Q B such thatAU A f B = f B .3.<strong>2.</strong> The category of balanced bimodule functors. We will construct here themonoidal category of balanced bimodule functors with respect to a monad.Definition 3.43. Let A = (A, m A , u A ) be a monad over a category A such thatAhas coequalizers and the underlying functor A preserves coequalizers. Let us definethe category ( A A ← A A) of balanced bimodule functors as followsOb Objects are functors A Q A : A A→ A A where Q : A → A is an A-A-bimodulefunctor such that Q A preserves coequalizers.M Morphisms are functorial morphisms A f A : A P A → A Q A where f : P → Q isa functorial morphism of A-A-bimodule functors.□
Proposition 3.44. Let A = (A, m A , u A ) be a monad on a category A such that theunderlying functor A preserves coequalizers and let A P A , A Q A ∈ Ob (( A A ← A A)).Then the functor A P AA Q A ∈ Ob (( A A ← A A)).Proof. We will prove that A P AA Q A = A (P AA Q) A . Let us consider the functorP AA Q : A → A. Since A Q is a right A-module functor by Proposition 3.31, thenalso P AA Q is a right A-module functor by Lemma 3.17. Thus, we can consider((P AA Q) A, p PAA Q) = Coequ Fun(µAPAA QAU, P AA Q A Uλ A)i.e.Lem3.17= Coequ Fun(PA µ A AQAU, P AA Q A Uλ A)P A preserves coequ= (P AA Q A , P A p A Q)(19) ((P AA Q) A, p PAA Q) = (P AA Q A , P A p A Q) .Now, observe that (P AA Q) Ais a left A-module functor by Proposition 3.30 andP AA Q A is a left A-module functor by Lemma 3.17. So we can consider both liftingfunctors : A (P AA Q) A and A (P A ) A Q A and we haveandHenceAU A ((P AA Q) A ) Pro3.31= (P AA Q) A(19)= P AA Q APro3.31= A U A (P A ) A Q AAUλ AA ((P AA Q) A) Pro3.30= A µ A (P AA Q) A(19)= µ PAA Q ALem3.17= A µ PA AQ APro3.30= A Uλ AA (P A ) A Q A .A ((P AA Q) A ) Pro3.31= A (P AA Q) A = A (P A ) A Q APro3.31= A P AA Q A .ThusAP AA Q A = A (P AA Q) Awhere P AA Q : A → A is an A-bimodule functor satisfying the required conditions.□Proposition 3.45. Let A = (A, m A , u A ) be a monad on a category A such that theunderlying functor A preserves coequalizers. Then A A A ∈ Ob (( A A ← A A)) and itis the unit element for the category ( A A ← A A).Proof. Since A is a monad, in particular an A-bimodule functor. Then we canconsider A A A ∈ Ob (( A A ← A A)) as the object coming from the endofunctor A :A → A. By definition we have(A A , p A ) = Coequ Fun (m AA U,A A Uλ A ) = ( A U, A Uλ A )and it is a left A-module functor by Proposition 3.13.considerAA A = A ( A U) = Id A A37By Lemma 3.29, we can
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: where A UG B F : B → A is such th
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 150 and 151:
150Now we compute(hQ) ◦ ( Qχ )
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
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Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...

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