Operator-Version of Schroeder-BernsteinDate: 11/13/97 at 15:46:32 From: Djordje Milicevic Subject: Operator-version of Schroeder-Bernstein? Here's the problem. It is well-known that by the Schroeder-Bernstein theorem, if there is an injection f:A -> B and a surjection g:A -> B, then there is also a bijection h:A -> B. My question relates to some algebraic structures, say groupoids (A,*) and (B,~). Suppose there is a monomorphism f:(A,*) -> (B,~) and an epimorphsim g:(A,*) -> (B,~). Does it follow that there necessarily exists an isomorphism h:(A,*) -> (B,~) ? If not, would it be true if we impose some more specific demands on * and ~ (for example, if we let (A,*) and (B~) be groups or whatever)? In fact, I even don't see the way to start out the regular proof of SBT (which would be to prove that there is a monomorphism from B to A). Thank you for your answer. I would be very grateful for any specific reference you can guide me to. Djordje Milicevic Date: 11/14/97 at 17:37:47 From: Doctor Ceeks Subject: Re: Operator-version of Schroeder-Bernstein? Hi, The answer to your first question is that it doesn't follow for groups. Here's a counter-example: Let D be a countable direct sum of copies of the integers Z. Let A = D direct sum Z/2Z. Let B = D direct sum Z/2Z direct sum Z/2Z. Then A and B are not isomorphic as groups (as can be seen by counting how many elements there are of finite order in the two groups). However, f:A\to B defined by sending Z/2Z to one of the factors Z/2Z in B and using any group isomorphism from D->D is an injective homorphism of groups. Also, g:A\to B defined by sending Z/2Z to one of the factors Z/2Z in B and sending all but one of the summands of D isomorphically to D, and the remaining one surjectively onto the other factor Z/2Z is a surjective group homomorphism. There definitely are, however, algebraic structures which you can impose on the category you are working in to make the theorem true, but I cannot think of any non-trivial ones immediately. Examples of trivial ones are working in the category of finite groups or the category of finite dimensional vector spaces. I think the result may be true for infinite dimensional vector spaces over Q too, because knowing the dimension plus the cardinality may determine the vector space up to isomorphism. That is, consider the category of Q-vector spaces with morphisms being Q-linear maps. Let A and B be Q-vector spaces and f, g as in your theorem. If A is finite dimensional, then the existence of f and g show that B is also finite dimensional and the same dimension. If A and B are not finite dimensional, from Schroeder-Bernstein, we know as sets, they have the same cardinality. It seems that the cardinality of a Q-vector space is equal to the cardinality of any basis. (A basis is a collection of vectors for which every vector in the vector space is a finite linear combination of vectors in the basis.) But two vector spaces with bases of the same cardinality are isomorphic. -Doctor Ceeks, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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