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**Example text**

Y : b - c ) such 1Ii:tt ( I ) Q 7 == / 3 c y (resp. y o a = yap), and (2) whenever a13 = PI3 (resp. d a = dfl) there is exactly one E with 6 = y E (resp. I3 = E y). 5. An equalizer is always a monomorphism. Prove and dualize. 0 0 Let a : a - + c and 8: b + c (resp. a : c - + a and 8: c + b) be two morphisms. ) of a and p is a couple of morphisms a’:d + b and p’: d -+ a (resp. a’: b + d and p’: a + d ) such that (1) ap’ = pa’ (resp. a‘p = p’a), and (2) whenever up‘‘ = pa’’ (resp. a “ p = 8 “ a ) there is exactly one y with a’y = a“ and p ’ y = /3” (resp.

3) (%, U ) has an extension in ($' 3, V ) iff there is a monotransforrnation p : U -+ U' such tha! (%, C7') is realizable in (23, V ) . 17. Proving a realizability of (%, U ) ir, (23, V ) is usually based on the following trivial observation: (a,U ) is realizable in (23,V ) iff there is a mapping F : obj21 --* obj23 such that (1) For all a E objELI, VF(a) = U(a), (2) a mapping f : U(a)-+ U(b) carries a morphism a -+ b iff it carries a morphism F(a) F(b). 18. ). In the sequel, it will be used without being mentioned, including the cases when used in combination with the facts that every realization is a strong embedding, every strong embedding is an extension, and every extension is a full embedding.

In 23. The i7, given by (E) For a set A denote by Q A the functor Set Set defined by Q A ( X )= X A , QA(f)(5) = f o 5. We have Qz E Q ; indeed, it suffices to define T : Q 2 -+ Q by = (t(O), {(l))(cf. l), similarly, Q1 E l,,,. (F) Denote by Pa the functor SetoP-+ Set defined by P A ( X )= A X , PA(f)(5) = 5 o f , by P - the functor defined by P - ( X ) = ( M M c X } , P-(f) ( M ) = f-'(M) (the inverse image of M under f). We have P2 E P - ; indeed, it suffices to define T : P2 -+ P - by ~ " (=l )c-'({l}).