A new characterization of injective and surjective functions and group homomorphisms ANDREI TIBERIU PANTEA* Abstract. A model of a function f between two non-empty sets is defined to be a factorization f ¼ i, where ¼ is a surjective function and i is an injective Æ ± function. In this note we shall prove that a function f is injective (respectively surjective) if and only if it has a final (respectively initial) model. A similar result, for groups, is also proven. Keywords: factorization of morphisms, injective/surjective maps, initial/final objects MSC: Primary 03E20; Secondary 20A99 1 Introduction and preliminary remarks In this paper we consider the sets taken into account to be non-empty and groups denoted differently to be disjoint. For basic concepts on sets and functions see [1]. It is a well known fact that any function f : A B can be written as the composition of a surjective function ! followed by an injective function. Indeed, f f 0 id , where f 0 : A Im(f ) is the restriction Æ ± B ! of f to Im(f ) is one such factorization. Moreover, it is an elementary exercise to prove that this factorization is unique up to an isomorphism: i.e. if f i ¼, i : A X , ¼: X B is 1 ¡ Æ ±¢ ! 1 ! another factorization for f , then we easily see that X i ¡ Im(f ) and that ¼ i ¡ f 0. The Æ Æ ± purpose of this note is to study the dual problem. Let f : A B be a function. We shall define ! a model of the function f to be a triple (X ,i,¼) made up of a set X , an injective function i : A X and a surjective function ¼: X B such that f ¼ i, that is, the following diagram ! ! Æ ± is commutative. A model (X ,i,¼) for the function f is called an initial model if for any other model (Y , j,p) of the function f there is a unique function g : Y X such that g j i and p g ¼.A ! ± Æ ± Æ model (X ,i,¼) for the function f is called a final model if for any other model (Y , j,p) of the function f there is a unique function g : X Y such that g i j and ¼ g p. ! ± Æ ± Æ *FACULTY OF AUTOMATIC CONTROL AND COMPUTERS,UNIVERSITY POLITEHNICAOF BUCHAREST,SPLAIUL INDEPENDEN ¸TEI 313, RO-060042 BUCHAREST 6, ROMANIAAND FACULTY OF MATHEMATICS AND COMPUTER SCIENCE,UNIVERSITYOF BUCHAREST,STR.ACADEMIEI 14, RO-010014 BUCHAREST 1, ROMANIA [email protected] 1 f A B i ¼ X Remark 1.1. For any function f : A B, there is at least one model. ! We may consider that A and B do not have any elements in common, otherwise, just consider a set B 0 B. Indeed, it is easy to see the remark is true if we consider X A B, Æ» Æ [ consider i to be the canonical injection and define ( f (a), if a A ¼(x) 2 . Æ b, if b B 2 2 Initial and final models of a function The next theorem gives a new characterization of injective and surjective maps. Theorem 2.1. Let f : A B be a function. Then: ! a) f has a final model if and only if f is injective. b) f has an initial model if and only if f is surjective. Proof. a) " " Let f be an injective function. In this case, (B, f ,id ) is obviously a model (Æ B for f and we will now show that it is a final model. Let (Y , j,p) be another model. We want there to be a unique function g : Y B such that g j f and id g p. The existence is ! ± Æ B ± Æ obvious since g p satisfies the relations. The uniqueness is given by the fact that id g p, Æ B ± Æ and so the unique function that satisfies this is g p. Æ " " Let f : A B, assume (X ,i,¼) is a final model of f and ® X . Suppose f is not Æ) ! 62 injective. Therefore, there are two elements of A, a a such that f (a ) f (a ). Let Y 1 6Æ 2 1 Æ 2 Æ X {®}. We define j : A Y and p : Y B such that [ ! ! ( ¼(x), if y X j(a) i(a) and p(x) 2 , Æ Æ f (a ), if y ® 1 Æ a A and y Y . 8 2 8 2 With this construction, (Y , j,p) is a model for f as a A, 8 2 (p j)(a) p ¡j (a)¢ p (i (a)) ¼(i (a)) f (a), ± Æ Æ Æ Æ since i(a) X and (X ,i,¼) is a model for f . We now define g ,g : Y X as 2 1 2 ! ( ( y, if y X y, if y X g1(y) 2 and g2(y) 2 . Æ i(a ), if y ® Æ i(a ), if y ® 1 Æ 2 Æ 2 Then g g , since i(a ) i(a ) because i is injective. We are now going to prove that g 1 6Æ 2 1 6Æ 2 1 and g2 are morphisms of models (i. e. the triangles are commutative) which contradicts the hypothesis that X is a final model (X is final g g ). Firstly, Æ) 1 Æ 2 ¡ ¢ ¡ ¢ (g1 j)(a) g1 j (a) g1 i(a) i(a), ± Æ Æ |{z} Æ X 2 a A, which means that g j i and, analogously, g j i. Now, for y Y we have 8 2 1 ± Æ 2 ± Æ 2 ( ( ¡ ¢ ¼(y), if y X ¼(y), if y X (¼ g1)(y) ¼ g1(y) 2 2 p(y) ± Æ Æ ¼(i(a )), if y ® Æ f (a ), if y ® Æ 1 Æ 1 Æ and, analogously, ( ( ¡ ¢ ¼(y), if y X ¼(y), if y X (¼ g2)(y) ¼ g2(y) 2 2 p(y). ± Æ Æ ¼(i(a )), if y ® Æ f (a ), if y ® Æ 2 Æ 2 Æ In conclusion, g g and both g g and g g commutatively close the diagram 1 6Æ 2 Æ 1 Æ 2 f A B i ¼ X j p g X {®} [ which contradicts the fact that (X ,i,¼) is a final model. The proof is now complete. b) " " Let f be an surjective function. In this case, (A,id , f ) is obviously a model for (Æ A f and we will now show that it is an initial model. Let (Y , j,p) be another model. We want there to be a unique function g : A Y such that id g j and p g f . The existence is ! A ± Æ ± Æ obvious since g j satisfies the relations. The uniqueness is given by the fact that id g j, Æ A ± Æ and so the unique function that satisfies this is g j. Æ " " Let f : A B, assume (X ,i,¼) is an initial model of f and { , } two different Æ) ! ¡1 1 elements which are not part of A,B or X . Suppose f is not surjective. Therefore, there exists ® X i(A) (otherwise, i is surjective, and so bijective and f ¼ i which implies that f 2 à Æ ± is a composition of surjective functions so it is surjective, contradiction). Let Y X {®} Æ à [ { , }. We define j : A Y and p : Y B such that ¡1 1 ! ! ( ¼(y), if y X {®} j(a) i(a) X and p(y) 2 à , Æ 2 Æ ¼(®), if y { , } 2 ¡1 1 a A and y Y . 8 2 8 2 3 Obviously, j is an injective function and p is surjective because p(Y ) ¼(X {®}) {¼(®)} ¼(X ) B. Æ à [ Æ Æ Moreover, (p g)(a) p ¡j(a)¢ p¡ i(a) ¢ ¼(i(a)) f (a), ± Æ Æ |{z} Æ Æ X {®} 2 à so (Y , j,p) is a model for f . We now define g ,g : X Y to be the functions 1 2 ! ( ( x, if x X {®} x, if x X {®} g1(x) 2 à and g2(x) 2 à . Æ , if x ® Æ , if x ® ¡1 Æ 1 Æ It is straightforward that g g , since . We shall now prove that g and g are 1 6Æ 2 1 6Æ ¡1 1 2 morphisms of models which contradicts the fact that (X ,i,¼) is an initial model (X is initial g g ). First of all, Æ) 1 Æ 2 ¡ ¢ (g1 i)(a) g1 i(a) i(a) j(a), a A, ± Æ |{z} Æ Æ 8 2 X {®} 2 à so g i j and, in an analogous way, g i j. Now, for p g p g ¼ we have to split 1 ± Æ 2 ± Æ ± 1 Æ ± 2 Æ our analysis into two parts: • If x X {®}, we have 2 à g (x) g (x) x p ¡g (x)¢ p ¡g (x)¢ p(x) ¼(x), x X {®}. 1 Æ 2 Æ Æ) 1 Æ 2 Æ Æ 8 2 à • If x ®, we have g (®) and g (®) , so Æ 1 Æ ¡1 2 Æ 1 p ¡g (®)¢ p( ) ¼(®) p( ) p ¡g (®)¢ 1 Æ ¡1 Æ Æ 1 Æ 2 which shows that p g p g ¼. ± 1 Æ ± 2 Æ In conclusion, g g and both g g and g g commutatively close the diagram 1 6Æ 2 Æ 1 Æ 2 f A B i ¼ X j p g X {®} { , } à [ ¡1 1 which contradicts the fact that (X ,i,¼) is an initial model. The proof is now complete. 4 3 Generalization to groups The term model used for sets can be generalized to groups in a natural way, by considering the functions to be group homomorphisms and the sets to be groups.