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Operator-Version of Schroeder-Bernstein


Date: 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/   
    
Associated Topics:
College Modern Algebra

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