Page 276 - Textos de Matemática Vol. 46

266 L. SOUSA
3. Main result
Before stating and proving the main result, we need some properties on nontrivial right adjoints over Set. (By nontrivial we mean that there is some C 2 C such that UC has at least two elements.) In particular, we will see that, for faithful nontrivial adjunctions (F,U,⌘,") : Set ! C, if f : X ! Y is a monomorphism of Set, then the square
X ⌘X //UFX (3.1) f UFf
✏✏ ⌘Y // ✏✏
Y UFY
is a pullback. In the terminology of [5], this means that every monomorphism f:X!Y issplitovertheidentitymorphismidY :Y !Y.
Part (a) of the following lemma is showed in Manes [10] (Proposition 5.2 and Proposition 5.42).
Lemma 3.1. Let (F,U,⌘,") : Set ! C be a nontrivial adjunction with U faithful. Then:
(a) The unit ⌘ is pointwise injective and F preserves monomorphisms.
(b) Every monomorphism f : X ! Y of Set is split over the identity mor-
phism idY .
Proof. (a) Given X 2 Set, and two di↵erent elements x,y 2 X, let C be an object of C such that UC has at least two elements, a and b. Define h : X ! UC byh(x)=aandh(z)=bforallz6=x.Nowleth# bethemorphisminCsuch that Uh# · ⌘X = h. Since h(x) 6= h(y), then ⌘X(x) 6= ⌘X(y).
Let now m : X ! Y be an injective map. If X 6= ;, then m is a split monomorphism, thus the same is true for Fm. If X = UFX = ;, UFm is a monomorphism since it has empty domain, and then Fm is a monomorphism since U being faithful reflects isomorphisms. If X = ; and UFX 6= ;, consider the diagram
; ⌘; //UF; << OO
m tUFm
✏✏ //✏✏
Ut# Y ⌘Y UFY
where t is any map from Y to UF;. Then we have U(t#Fm)⌘; = Ut#·⌘Y ·m = tm = ⌘;. Thus t#Fm = idF;, so Fm is a monomorphism.