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

14(2) Hom B (−, ηY ′ ) is a functorial isomorphism(3) ηY ′ is an isomorphism (η is a functorial isomorphism).Proof. Since (L, R) is an adjunction, a X,Y : Hom A (LY, X) → Hom B (Y, RX) is anisomorphism for every X ∈ A and for every Y ∈ B, so that (1) is equivalent to (2).(3) ⇒ (2) Let η −1 Y ′ be the two-sided inverse of ηY ′ . Then Hom B (−, η −1 Y ′ ) is theinverse of the functor Hom B (−, ηY ′ ) . In fact, let f ∈ Hom B (Y, Y ′ ) and compute[ (HomB −, η −1 Y ′) ◦ Hom B (−, ηY ′ ) ] ((f) = Hom B −, η −1 Y ′) (ηY ′ ◦ f)= ( η −1 Y ′) ◦ (ηY ′ ) ◦ f = fand[HomB(−, ηY ′ ) ◦ Hom B(−, η −1 Y ′)] (f) = Hom B (−, ηY ′ ) (( η −1 Y ′) ◦ f )= (ηY ′ ) ◦ ( η −1 Y ′) ◦ f = f.Thus Hom B (−, ηY ′ ) is a functorial isomorphism.(2) ⇒ (3) Since Hom B (−, ηY ′ ) is a functorial isomorphism, in particularHom B (RLY ′ , ηY ′ ) : Hom B (RLY ′ , Y ′ ) → Hom B (RLY ′ , RLY ′ ) is an isomorphism.Thus, there exists f ∈ Hom B (RLY ′ , Y ′ ) such that (ηY ′ ) ◦ f = Id RLY ′ which impliesthat ηY ′ is an epimorphism. Moreover we also have Hom B (Y ′ , ηY ′ ) (Id Y ′) = ηY ′ =(ηY ′ ) ◦ f ◦ (ηY ′ ) = Hom B (Y ′ , ηY ′ ) (f ◦ (ηY ′ )) . Since Hom B (−, ηY ′ ) is a functorialisomorphism, also Hom B (Y ′ , ηY ′ ) is an isomorphism. Thus we deduce that Id Y ′ =f ◦ (ηY ′ ) which implies that ηY ′ is also a monomorphism and moreover ηY ′ has atwo-sided inverse f : RLY ′ → Y ′ .□Remark <strong>2.</strong>29. Note that, for every f ∈ Hom B (Y, Y ′ ) we haveL Y,Y ′ (f) = [ a −1LY ′ ,Y ◦ Hom B (Y, ηY ′ ) ] (f) = a −1LY ′ ,Y (ηY ′ ◦ f)= (ɛLY ′ ) ◦ (LηY ′ ) ◦ (Lf) (L,R)adj= Lf.Lemma <strong>2.</strong>30. Let (L, R) be an adjunction with unit η and counit ɛ, where L : B → Aand R : A → B. For every X ∈ A the following conditions are equivalent:(1) R X,− = a −,RX ◦ Hom A (ɛX, −) is a functorial isomorphism(2) Hom A (ɛX, −) is a functorial isomorphism(3) ɛX is an isomorphism (ɛ is a functorial isomorphism).Proof. Dual to proof of Lemma <strong>2.</strong>28.Remark <strong>2.</strong>3<strong>1.</strong> Note that, for every f ∈ Hom A (X, X ′ ) we haveR X,X ′ (f) = [a X ′ ,RX ◦ Hom A (ɛX, X ′ )] (f) = a X ′ ,RX (f ◦ ɛX) = R (f ◦ ɛX) ◦ (ηRX)= (Rf) ◦ (RɛX) ◦ (ηRX) (L,R)adj= Rf.Proposition <strong>2.</strong>3<strong>2.</strong> Let (L, R) be an adjunction with unit η and counit ɛ, whereL : B → A and R : A → B. Then R is full and faithful if and only if ɛ is afunctorial isomorphism.Proof. To be full and faithful for R means that the mapφ : Hom A (X, X ′ ) −→ Hom B (RX, RX ′ )□