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The Cantor-Bernstein theorem in the category of sets (A injects in B, B injects in A => A, B equivalent) holds in other categories such as vector spaces, compact metric spaces, Noetherian topological spaces of finite dimension, and well-ordered sets.

However, it fails in other categories: topological spaces, groups, rings, fields, graphs, posets, etc.

Can we caracterize Cantor-Bernsteiness in terms of other categorical properties?

[Edit: Corrected misspelling of Bernstein]

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I usually hear this referred to as the Schroeder-Bernstein theorem. Regardless of the Cantor/Schroeder switch, I think you're misspelling Bernstein. –  Scott Morrison Oct 18 '09 at 20:59
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Ooops for the mispelling. As for the denomination of the theorem, I got that from my lecturer, Peter Johnstone. If I recall correctly, he claimed that Cantor was actually the first to correctly prove the theorem, and that Schroeder got his proof wrong. –  Randomblue Oct 18 '09 at 21:20
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OTOH, in any category, if there are monomorphisms $A \hookrightarrow B$ and $B\hookrightarrow A$, then for any $X$ the sets $\operatorname{Hom}(X,A)$ and $\operatorname{Hom}(X,B)$ are isomorphic, by unpacking the word "monomorphism" and applying Schroeder-Cantor-Bernstein (one of those rare results with non-alphabetized names). If this isomorphism could be made natural in $X$, then by Yoneda we would have an isomorphism $A\cong B$. But of course it cannot, in general. –  Theo Johnson-Freyd Jul 19 '10 at 4:15
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I think you're still misspelling Cantor-Bernsteinness. –  Tom Goodwillie Sep 21 '10 at 19:01
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Maybe it is interesting to mention that this thread was mentioned in the book Proofs of the Cantor-Bernstein Theorem: A Mathematical Excursion by Arie Hinkis, see footnote on page 397. –  Martin Sleziak May 19 at 11:42

5 Answers 5

up vote 47 down vote accepted

Whenever the objects in your category can be classified by a bounded collection of cardinal invariants, then you should expect to have the Schroeder-Bernstein property.

For example, vector spaces (over some fixed field K) or algebraically closed fields (of some fixed characteristic) can each be classified by a single cardinal invariant: the dimension of the vector space, or the transcendence degree of the field.

More interesting example: countable abelian torsion groups. Suppose A and B are two such groups, A is a direct summand of B, and vice-versa; are they isomorphic? By Ulm's Theorem, A and B are determined up to isomorphism by countable sequences of cardinal numbers -- namely, the number of summands of Z_p^infty and the "Ulm invariants," which are dimensions of some vector spaces associated with A and B. All of these invariants behave nicely with respect to direct sum decompositions, so it follows that A and B are isomorphic. (See Kaplansky's Infinite Abelian Groups for a very nice, and elementary, proof of all this.)

If you like model theory, I could tell you a lot about when the categories of models of a complete theory have the Schroeder-Bernstein property (under elementary embeddings). If not, at least I can tell you this:

  1. Categories of structures with "definable" partial orderings with infinite chains (e.g. real-closed fields, atomless Boolean algebras) will NOT have the S-B property. Again, I need some model theory to make this statement precise...

  2. Let C be a first-order axiomatizable class of structures (in a countable language) which is "categorical in 2^{aleph_0}" -- i.e. any two structures in C of size continuum are isomorphic. Then C has the S-B property with respect to elementary embeddings. (This generalizes the cases of vector spaces and algebraically closed fields.)

Addendum: A completely different way that a category C might be Schroeder-Bernstein is if every object is "surjunctive" (i.e. any injective self-morphism of an object is necessarily surjective). This covers Justin's example of the category of well-orderings.

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There are monomorphisms of countable abelian torsion groups that are not direct summands, e.g. Z/3 into Z/9. I'm not much of a group theorist, and I don't know Ulm's Theorem. Is it true that monomorphisms in both ways implies an isomorphism, or just split monomorphisms (direct summands)? –  Theo Johnson-Freyd Jul 19 '10 at 4:11
    
@Theo: no, having monomorphisms $f: G \rightarrow H$ and $g: H \rightarrow G$ both ways does not imply that two countable abelian torsion groups are isomorphic. For instance, you can take $G$ to be the direct sum of a countably infinite number of copies of $Z / {9 Z}$ and take $H$ to be the direct sum of $G$ and a copy of $Z / {3 Z}$. –  John Goodrick Jul 21 '10 at 4:29
    
How does this categorical argument work in the case of computable sets? (Myhill isomorphism theorem) –  Diego de Estrada Oct 3 '10 at 12:42
    
@Diego: Offhand, I don't know of a "categorical" proof of the Myhill isomorphism theorem, whatever this would mean (I'm not saying that there isn't one, but I don't know of one). To be clear, I didn't mean that "classifiability by a bounded set of cardinal invariants" is necessary to have S-B in your category, just that it is (with a suitable interpretation of the terms) a sufficient condition for S-B. –  John Goodrick Oct 4 '10 at 18:16
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@JohnGoodrick This is a very interesting answer. Is there a good place where the slogan of the very first sentence is given precise expression? –  Todd Trimble Feb 16 at 0:46

We discussed this at the Secret Blogging Seminar.

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Thank you for the link. However, the discussion seems to be mostly case by case analysis, not "high level" category theory. –  Randomblue Oct 18 '09 at 21:29
    
A new version of this question I liked: the category of factors, the category of II_1 factors. –  Noah Snyder Oct 18 '09 at 23:29

Let me point out a curious (non-categorical) twist to an elementary observation. The elementary observation is that the Cantor-Schroeder-Bernstein property fails for linear orderings, even if we require the injections to map onto an initial segment. That is, you can have two non-isomorphic linear orders, each isomorphic to an initial segment of the other. One example is the same as a familiar example for the topological case, the closed interval [0,1] and the half-open interval [0,1) of real numbers. And of course, the situation is the same for final segments. The curious twist is that, if a linear order A is isomorphic to an initial segment of another linear order B, while B is isomorphic to a final segment of A, then A and B are isomorphic.

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Here's another example where a type of Cantor-Schroeder-Bernstein theorem holds. In this paper of R. Bumby, it was proved that if $A$ and $B$ are injective modules over a ring that can be embedded in each other, then $A \cong B$.

An immediate corollary is that if any two modules over a ring embed in one another, then their injective hulls are isomorphic.

I wonder whether this fits into John Goodrick's answer above. I'm not aware of any bounded collection of cardinal invariants that classify injectives over an arbitrary ring. (If the ring is right noetherian then each injective decomposes into a direct sum of indecomposable injectives, and this allows us to classify the injectives. But Bumby's result holds for arbitrary rings!)

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Here is another example where a type of Cantor-Schroeder-Bernstein theorem holds: the category of compact metric spaces, which is in fact surjunctive according to the terminology described by John Goodrick. This remark is motivated by the question

Isometries between metric spaces

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