[FOM] Bourbaki's general theory of isomorphism‏

Victor Makarov viktormakarov at hotmail.com
Fri Jun 25 14:07:03 EDT 2010

 In reply to Oran Magal (Thu 6/24/10 7:50 PM):

When we state "A is isomorphic to B", we tacitly assume, that A and B are models of some theory T.

We can further assume that the theory T is an extension of ZFC by adding to ZFC (or to an extension of ZFC) some new constants and axioms "defining" the new constants.
For example, the group theory can be introduced in the following way:

group := {G, m | A };

where G, m are , respectively, the set on which the group is defined, and the law of composition, A is a conjuction of axioms of the group theory.

Any pair (X, h) such that the formuala A(X,h) is a theorem, is a concrete group (or a model of the group theory).
 A(X,h) denotes the result of substitution in the formula A, instead of the constants G, m, the terms X, h respectively. 

Let A = (X,h), A1 = (X1,h1) be some concrete groups. A is isomorphic to A1 iff there exists a bijection f from X to X1 such that f'(h) = h1, where f' is the canonical extension of f. 

Suppose, we have defined in the group theory some notion Q(G,m).  
That is, Q is a formula (or a term), possibly containing the constants G,m.
If A is a concrete group (X,h), let us denote A.Q the formula Q(X,h).

If the formuala A.Q is a theorem, then we say that Q is a property of A.
The intuitively obvious statement "Isomorphic groups have the essentially same properties" can be more precisely  expressed in the following way:
If the groups A, B are isomorphic then the formula A.Q is equivalent to the formula B.Q.
The formula Q must be transportable, i.e Q can not depend on some specific properties of the constants G,m.
The formula expressing "the empty set is an element of G" is an example of untransportable formula. Obviously, this formula is not preserved under isomorphisms.

Victor Makarov  		 	   		  

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