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The Cantor-Bernstein-Schroeder theorem states that if there are two sets $A$ and $B$ such that there exist injective (alternatively, surjective, assuming choice I think) maps $A \to B$ and $B \to A$, then $A$ and $B$ are in bijection. It then seems natural to try to strengthen the result to other structures. If $A$ and $B$ are, say, groups (or rings, modules, etc) such that there are injective homomorphisms from each to the other, then $A$ and $B$ are isomorphic.

I'm wondering if there are any results on this, or if there are known counterexamples. The theorem for sets alone is nontrivial, so I feel like any results about how it holds for other structures would be quite interesting.

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Just as a quick counterexample to "groups, rings, modules, etc.": if $G$ is the direct product of countably many copies of $\mathbb{Z}/4\mathbb{Z}$, then there is a natural injection $G \to \mathbb{Z}/2\mathbb{Z} \times G$ via inclusion in the second coordinate, and there is an injection $\mathbb{Z}/2\mathbb{Z} \times G \to G$ via $1 \mapsto 2$ in the first coordinate and identity in all other coordinates. –  Ben Dec 13 '12 at 4:33
    
Interesting, thanks. Both those comments together pretty much completely answer my question. I guess I'll accept if either of you posts them as an answer. –  Carl Dec 13 '12 at 5:48
    
(I've posted the comment as an answer.) –  Andres Caicedo Dec 13 '12 at 6:40
    
I think it interesting that this theorem can be proven without using the axiom of infinity. I conjecture that it's possible to prove the theorem for proper classes $A$ and $B$, but only by accepting the axiom of infinity. –  dfeuer Apr 2 '13 at 22:19

1 Answer 1

up vote 11 down vote accepted

This is an interesting topic, and has been studied in many contexts. See this question in MO for examples, counterexamples, and some links to the literature (far from exhaustive). In the context of model theory, John Goodrick (the author of the answer in the link above) has several papers on this topic, including his PhD dissertation.

One typically refers to these analogues as the (Cantor)-Bernstein-Schröder property: Given a class $C$ of objects and a class $\mathcal F$ of maps between them, the Bernstein-Schröder property for $C,\mathcal F$ means that if $X,Y$ are objects in $C$ and $f:X\to Y$ and $g:Y\to X$ are injections in $\mathcal F$, then there is a bijection in $\mathcal F$ between $X$ and $Y$. Typically, $C$ and $\mathcal F$ form a category, and the sought-after bijections would be equivalences in the category. This reformulation suggests further generalizations.

Here are some examples:

  • The class $C$ of well-orderings, with $\mathcal F$ the class of order-preserving maps.

  • The class of Polish spaces, with Borel maps.

  • (Assuming the axiom of choice.) The class of vector spaces (over a fixed field $K$), with $K$-linear maps. (Let me prove this, as the argument is simple enough. Note that if there is a linear injection $T:V\to W$, then any linearly independent set in $V$ is mapped to a linearly independent set in $W$. In particular, if $V$ and $W$ have a basis (as is the case under choice), then the dimension of $V$ is less than or equal than that of $W$. Hence, if there are linear injections $T:V\to W$ and $S:W\to V$, then $V$ and $W$ have the same dimension, and any bijection between a basis of $V$ and one of $W$ uniquely extends to a $K$-isomorphism between the spaces.)

The category of groups with group homomorphisms is an example of a category without the Bernstein-Schröder property. For example, consider $G=\bigoplus_{n\in\omega}\mathbb Z/4\mathbb Z$ and $H=G\oplus \mathbb Z/2\mathbb Z$, where $\oplus$ denotes direct sum. There are monomorphisms from $G$ into $H$ and from $H$ into $G$, but $G$ and $H$ are not isomorphic. (This is actually a typical example. I just noticed—after typing it—that this is precisely the example given in a comment above.)

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Thanks. I guess one other interesting question would be: Are there situations in ZFC where monomorphisms $A \to B$ and $B \to A$ are sufficient to get an isomorphism but epimorphisms are not, or visa-versa? –  Carl Dec 13 '12 at 6:51
    
Do we know about the category of fields? –  mez May 23 at 17:21

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