B' G(r B (x)) Defi n e t' (A) : [G(A), One e a s i l y v e r i f i e s t' = p(t), then By assumption, [[G-, Put B], that the m a p p i n g B']] b = k~(t)), is uni q u e w i t h then [G-, b'] This completes B'] is n a t u r a l by t' (A) (x) = t(x) .

M h PROPOSITION is also sup-complete CASE I. consisting of A O A Let CASE 2. A_ be inf-complete. _A A contains two cases: subcategory Ao, and arbitrary sums a generator A. Moreover, Ao, and arbitrary every object in A - - has representative Proof. sets of subobjects For any set X, let the direct sum of copies of x ~ X. Then G : Ens ~ A right adjoint = A__ A. 2. % = ~xeX = [ x~X Ao denote one for each element is a functor, which has as a the so-called " A___~ Ens. Ao, G(X) and quotient objects.

This property; = t', h e n c e for if also b' = A(t') = b. the first p a r t of the proof. b 'x = t(x), - 30Conversely, We w i l l assume Let B], [G-, B']] and onto. t' [G(FB(X)), : [G -, B] ~ B], we h a v e easily verifies that there exists Thus t' = [G -,b], t' = [G -, b']. [G -, B']. Then, this b is n a t u r a l : B ~ in B' such the m a p p i n g by uniqueness. B e f o r e we can a s s e r t One is onto. for all This c o m p l e t e s ~ B~ x. By a s s u m p t i o n , is one-one, bx = b'x x that and so our m a p p i n g Then for any t'~FB(X) ) : G(FB(X) ) a unique To see that = b [B, B']-~ [[G-, s h o w that the m a p p i n g is o n e - o n e b' sup(Got B) = (S, I) that assume x that a l s o : G(A) ~ B, h e n c e the proof.