Talk:Surjection iff Right Inverse

The statement that any surjection has a right inverse is in fact equivalent to the axiom of choice. Is there a page about this? I can easily provide three equivalent forms of the axiom of choice, and a proof of equivalence:


 * Any surjection $f: X \to Y$ has a right inverse.


 * For every $X\neq \emptyset$ there exists a function:
 * ${\displaystyle f: \mathcal{P}\left(X\right)\setminus\left\{\emptyset\right\} \to X}$

such that $f(Y)\in Y$ for each $Y\subset X$ with $Y\neq \emptyset$.


 * Let $\left(X_i\right)_{i\in I}$ be a family of sets such that $X_i \neq \emptyset$ for each $i\in I$.
 * Then:
 * ${\displaystyle \prod_{i\in I} X_i \neq \emptyset}$,
 * where
 * $\prod_{i\in I} X_i = \left\{ f: I \to \bigcup_{i\in I} X_i : f(i) \in X_i \ \forall i\in I\right\}. $

Who has ideas on how to add them? I do not want to add it while there are other proofs of equivalence on, say, two out of these.

Second of all, I have found plenty more equivalent forms of the axiom of choice, while these three easily follow from on another, the others might not... JSchoone 16:31, 31 January 2012 (EST)


 * Excuse me ... you have actually read this page?? There are two proofs, both of which are specific about the use of AoC. There is even a category for proofs of AoC and there are already quite a few equivalent statements. AoC itself is defined as your final statement, using the language of cartesian products. --prime mover 17:23, 31 January 2012 (EST)


 * I have, but it says nowhere, that it is equivalent to the axiom of choice, it just says that this page follows from the axiom of choice. It is quite a difference. Still I cannot find the page where this is proven, so if you can, please tell me where it is and accept my apologies for not looking well enough. JSchoone 05:46, 1 February 2012 (EST)


 * Sorry, I misunderstood. I understand what you mean now.
 * What we are saying here, then, is:
 * a) If the Axiom of Choice holds, then Surjection iff Right Inverse is true.
 * b) If the Axiom of Choice does not hold, then there exists a set whose subsets do not admit a choice function. Therefore it is possible to construct a surjection such that the preimages of the elements of the range do not admit a choice function, and therefore such a right inverse can not be constructed.
 * Fair enough, it should be straightforward to amend this page so that (b) above can be formulated. Feel free to do so.


 * What puzzled me was your two statements of the Axiom of Choice above, as though it were new to ProofWiki. As I said,. this is already all covered, except for (b) above, as I mentioned. We also have several other equivalences of AoC. The usual wording is: this result is true if AoC holds - but we rarely add the wording in the other direction: "if AoC fails to hold, then so does this result." This is what is usually meant when we say "This result depends on AoC." I never thought of is as worth raising. --prime mover 08:57, 1 February 2012 (EST)


 * The converse statement appears more useful and powerful if one states it as: 'This result implies AoC'. The non-emptiness of Cartesian products and the principle of cardinal comparability (trichotomy law) are too important for me to consider the negation of AC; more so as negating AC makes stuff (like Banach-Tarski) only undecidable, not false. --Lord_Farin 09:44, 1 February 2012 (EST)