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Given a family of sets $(A_i : i \in I)$, we define the disjoint union: $$\sum_{i \in I} A_i = \bigcup_{i \in I} (\{i\} \times A_i).$$

There is a surjection $\sum_{i \in I}A_i \to \bigcup_{i \in I} A_i$ given by $(i,a) \mapsto a$, so by the Axiom of Choice there is an injection $\bigcup_{i \in I} A_i \to \sum_{i \in I}A_i$.

My question is whether AC (or some fragment of it) is required to prove that there is an injection $\bigcup_{i \in I} A_i \to \sum_{i \in I}A_i$.

(If I remember correctly, it is not known whether "every surjective image of any set $X$ injects into $X$" implies AC.)

Asaf Karagila
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Trevor Wilson
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    Well, isnt't an injection essentially a choice function? Oh, wait - only if it is a left invers to the porojection, but you donÄt require that?! – Hagen von Eitzen Mar 06 '13 at 21:12
  • Are you talking about the last sentence? If so, it sounds you have already identified the issue there. – Trevor Wilson Mar 06 '13 at 21:15
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    AC is equivalent to every surjective function having a right inverse. – JSchlather Mar 06 '13 at 21:16
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    If $I$ is an ordinal, we have $a\mapsto\langle\min{i\in I\mid a\in A_i},a\rangle$, which is even inverse to the projection and shows we at least don't need countable choice for the countable case. – Hagen von Eitzen Mar 06 '13 at 21:17
  • @Hagen Yes, if you can choose an element of $I$ for each element of the union then that is sufficient. – Trevor Wilson Mar 06 '13 at 21:20

1 Answers1

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The assertion "If $f\colon X\to A$ is surjective then there is $g\colon A\to X$ injective" is known as The Partition Principle.

Indeed it is still open whether or not the partition principle implies the axiom of choice. (It is one of my goals for the foreseeable future, to solve this problem, although it's a goal I am fully aware I am unlikely to achieve.)

To see that the formulation $\bigcup A_i\leq\sum A_i$ is equivalent to the Partition Principle, one direction is trivial (clearly PP implies this thing), and in the other direction suppose $f\colon A\to B$ is surjective, take $I=A$ and $X_a=\{f(a)\}$. We have that $\bigcup X_a=B$ while $\sum X_a=f$ which is naturally in bijection with $A$. Requiring, if so, an injection from the union into the sum implies an injection from $B$ back into $A$, as wanted.

The assertion "every surjection splits" indeed implies the axiom of choice, that is to say the map $(i,a)\mapsto i$ is a surjection then there is an injection which splits it, is exactly asserting a choice function.

But requiring only the existence of at least one injection whenever there is a surjection is not enough to ensure every surjection splits, at least not as we know it. It goes even further, I don't think that we know that many counterexamples too.

The only sets I am aware of that have the property that whenever $f\colon X\to Y$ is surjective then there is $g\colon Y\to X$ injective are sets whose surjections split, and those are strong $\kappa$-amorphous sets, that is sets that every partition into two sets implies one is smaller than $\kappa$; and that every partition into no less than $\kappa$ parts is almost all singletons (almost all: meaning all but $<\kappa$ many parts).

For example, if $A$ is strongly amorphous (which means that it cannot be split into two infinite sets, and every infinite partition is all but finitely many singletons) then every surjection from $A$ splits.

I am not aware of any sets other than those which have the partition principle in ZF, or even consistently having partition properties.

Asaf Karagila
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  • Is every surjection equivalent to one of the form that is mentioned in the question? I don't see why the partition property is necessary to find an injection in that particular case. – Trevor Wilson Mar 07 '13 at 00:10
  • @Trevor, Hm. I see your point. If $A_i$ were disjoint to begin with then the surjection is really a bijection. However in the general case I do believe that a reduction to PP is possible. I'll think about it some more. – Asaf Karagila Mar 07 '13 at 00:52
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    Thanks. I'm surprised that I was able to ask a question about AC that you didn't immediately know the answer to :) It came up in my class (I asked the class to prove it using AC) and so naturally I was curious about how much choice was needed. – Trevor Wilson Mar 07 '13 at 06:02
  • Well, it is easy to come up with hard AC questions. :-) Also PP is exactly what you wrote in that remark in the parenthesis. – Asaf Karagila Mar 07 '13 at 06:23
  • Yes, I learned about it from one of your posts here, but I forgot the name. – Trevor Wilson Mar 07 '13 at 06:33
  • Also, do you ever sleep? – Trevor Wilson Mar 07 '13 at 06:36
  • Well, I have been known to doze off from time to time. :-) – Asaf Karagila Mar 07 '13 at 06:51
  • Yeah, I told you it seemed related to PP. The proof is kinda silly because the family I use is a family of singletons. :-) – Asaf Karagila Mar 07 '13 at 07:03
  • Nice! I hoped there might be a simple trick like that, but that is even simpler than I thought – Trevor Wilson Mar 07 '13 at 07:08
  • Ditto. I just wrote it and it came out. I had to verify it three times before I was satisfied by the complete lack of complexity! :-) – Asaf Karagila Mar 07 '13 at 07:09