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This question might be silly.

I was wondering wether some form of choice is needed to be on firm footing when talking about the set of equivalence classes of certain things. For instance, when one talks of the set of equivalence classes of extensions of one $R$-module $N$ by another, $M$, $$\lbrace E\text{ such that there exists a short exact sequence }0\to N\to E\to M\to 0\rbrace/\sim$$ where two short exact sequences involving $E$ and $E'$ are equivalent if there is a commutative diagram $$\begin{array}{c} 0\to &N&\to E\to &M&\to 0\\ &\Vert&~~~\downarrow\phi&\Vert&\\ 0\to &N&\to E'\to &M&\to 0. \end{array}$$ In order to justify that this is indeed a set, I can only think of the following approach: take a section (a map of sets) $\sigma:M\to E$ of $p:E\to M$ and use this to obtain a bijection of sets $$E\simeq N\times M,~e\mapsto(n(e),p(e))$$ where $n(e)\in N$ is defined by $n(e)+\sigma(p(e))=e$, and argue that there is only a set of $R$-module structures on the set $N\times M$, so that in the above we may choose $E$ to always be $N\times M$ as a set, endowed with different $R$-module structures. If I'm not mistaken, the existence of sections to surjective maps depends on the axiom of choice.

A similar problem arises when one talks about the set of isomorphism classes of all $n$-dimensional vector bundles over a given space $X$, or that of principal $G$-bundles, although in this particular case I think one can directly argue over the set of all open covers of $X$.

I guess the question boils down to the following: is choice needed to show that given a map $X\to Y$ with fibers of constant cardinality, the cardinality of $X$ depends only on that of $Y$ and of the fiber?

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The collection you describe is not literally a set, however. The standard way to avoid the axiom of choice for these things is to use Scott's trick. –  Zhen Lin May 30 '13 at 19:00
    
I like the title... the idea of a general relation between classes (like class of objects or arrows) and sets –  user39158 Aug 9 at 18:03

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First of all, note that since the equivalence classes themselves are not sets, the collection of all equivalence classes is not a set either, it's not even a class. The correct phrasing is to say that there exists a set of representatives, rather than a class of representatives (which might not exist either, by the way). Of course if we want to use Scott's trick then the equivalence classes are cut down to be sets, and then it's a legit thing to say that there is only a set of equivalence classes.

Secondly, if we require that the exact sequences exist, and that the equivalence relation is defined by the existence of isomorphism between $E$ and $E'$. The equivalence classes themselves are well-defined without any appeal to the axiom of choice.

I believe, however, without sitting to verify the details I think it should be consistent to have a proper class of sets (of distinct cardinality) which are the countable union of pairs. The existence of an exact short sequence means that a particular pair is mapped onto $0$. So a proper class of sets can exist which have this sort of sequence, without the existence of an isomorphism, and therefore even using the set theoretical tricks to turns equivalence classes into sets, this is still going to be a proper class.

But that been said, if we want concrete sets $M$ and $N$, then in order to assure that the sections exist we need only the axiom of choice for families of size $|Y|$ whose members are the "constant" cardinality (we can just as well take $|X|$ to be sure, if we want to be sure of this).

If we assume such choice, then all these $E$ (as in the sequences) are isomorphic to subsets of $\mathcal P(N\times M)$, so there can only be a set of those.

For example if $Y$ is countable, then the axiom of countable choice is sufficient, if we also have that the fibers are all finite - or even pairs - then the axiom of choice for countable families of pairs should be sufficient. Usually saying that there is a choice function for finite sets, or for families of pairs, would be a nicer way of saying that (although stronger than required).

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Of course, in a case like this, we can also get around the collection of such extensions not being a set by noting that any such extension can be realized as a subset of a suitably chosen set. –  Tobias Kildetoft May 30 '13 at 19:09
    
Thanks. There is a little typo when you talk about the projective and free abelian groups. –  Olivier Bégassat May 30 '13 at 19:10
    
@Olivier: What typo? –  Asaf Karagila May 30 '13 at 19:12
    
A free abelian group is always projective, but it's the converse that requires choice (and apparantly is equivalent to it). –  Olivier Bégassat May 30 '13 at 19:13
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