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Suppose $f:A\to B$ is surjective. By definition for all $b\in B$, there is $a_b\in A$ s.t. $b=f(a_b)$. Define $g:B\to A$ by $$g(b)=a_b.$$ Clearly, $g$ is a well defined function, because $g(b)$ has no two different value, and $g(b)\in A$ for all $b$. Therefore, for all $b\in B$, $$f(g(b))=f(a_b)=b,$$ and thus $g$ is invertible at right. Where did I used Axiom of choice ?

user380364
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2 Answers2

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You are misleading yourself when you call this $a$ by $a_b$. You're implicitly assuming that this is uniformly identifiable. It is not. Instead, you should think as follows:

A function $f\colon A\to B$ is surjective, if for every $b\in B$ there is some $a\in A$ such that $f(a)=b$.

Can you define $g$ now? $g(b)=a$? What does it even mean? Which $a$? You'd retort with "Well, the one that is sent to $b$ of course". But what if there are many of them? Which one is "the one"?

For example, what is "the enumeration" of a countable set of reals? Clearly each countable set of reals can be enumerated, it's literally the definition of being countable. But which one is "the enumeration"? Or, given $r\in\Bbb R$, which one is the real number $r'$ such that $r-r'$ is rational, and $r'$ is the same for any $r+q$ when $q\in\Bbb Q$?

It's those instances of switching quantifiers where the axiom of choice lies. You need the axiom of choice to prove that there is a uniform way of choosing this enumeration, this $r'$, this $g(b)$. And while you might not quite know which one you ended up with, the rules of logic do let you assign it with a name: $g(b)$.

Additional reading:

  1. Axiom of Choice and Right Inverse

  2. Show that $f$ is right invertible.

  3. Axiom of Choice: Where does my argument for proving the axiom of choice fail? Help me understand why this is an axiom, and not a theorem.

Asaf Karagila
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  • There's probably a few other threads related to this topic. – Asaf Karagila Aug 30 '18 at 13:52
  • First of all, thanks a lot for your answer (and also, congrat' for your new position as moderator, I really appreciate the way you do your work). I see a little bit better (the example with an enumeration of $\mathbb Q$ is very enlightening). But also, what do you mean by "the enumeration" ? It's in fact what I don't really get, since there is an enumeration, so just talk one among the enumeration that exist... Same with surjectivity : if there are many $a$ s.t. $f(a)=b$, which one I'd talk ? Just one of them... why do I need :"the one"? (what in fact "the one" mean since there are several ?) – user380364 Aug 30 '18 at 14:00
  • It's not about $\Bbb Q$. It's about enumerating all the countable subsets of $\Bbb R$. Or in other words, consider the function that maps an enumerating function to its enumerated set. This is a surjection between two sets. What is the right inverse, then? – Asaf Karagila Aug 30 '18 at 14:01
  • Sorry but I'm really confuse now. What is an enumerating function ? But let go back to the surjective function. I think it's better. Consider $f:\mathbb R\to {0,1}$ a surjective function. I know that $f^{-1}(0)$ and $f^{-1}(1)$ are non empty. Take $x\in f^{-1}(0)$ and $y\in f^{-1}(1)$ (such element exist). Then define $g:{0,1}\to \mathbb R$ s.t. $g(0)=x$ and $g(1)=y$. This is correct, right ? No need choice axiom, does it ? It's really when you say : the right element or "the one" that confuse me. – user380364 Aug 30 '18 at 15:04
  • But you are thinking about choosing just two. Try infinitely many. An enumerating function is just an injection from the natural numbers into the reals. – Asaf Karagila Aug 30 '18 at 15:26
  • Ok perfect. So the technique in my previous comment is true. Let generalize a bit. So, let $f:\mathbb R\to \mathbb N$ a surjective function. I know that $f^{-1}(1)$ are non empty. I denote $x_i$ an element of $x_i\in f^{-1}(i)$. Set $g:\mathbb N\to \mathbb R$ defined by $g(i)=x_i$. Then $f\circ g=id_{\mathbb N}$. This is correct, right ? And I didn't use axiom of choice, did I ? – user380364 Aug 30 '18 at 15:36
  • What is this $x_i$ you speak of? Didn't you choose it? How did you choose infinitely many of these $x_i$s in a finite proof? – Asaf Karagila Aug 30 '18 at 15:46
  • In the same way that I chose $x$ and $y$ in my example with $f:\mathbb R\to {0,1}$. Why it was correct in this previous example, and not here with $f:\mathbb R\to \mathbb N$ ? (sorry to annoy you, but I really want to understand this subtlety :)) – user380364 Aug 30 '18 at 15:48
  • Have you read https://math.stackexchange.com/a/1839929/622 or the answers to the other links? – Asaf Karagila Aug 30 '18 at 15:54
  • Of course I read them (with attention). But it's really not clear why $\exists s\in S$ and Let $s$ an element of $S$ is not the same thing... I think the subtlety come from this fact... but that's why I'm trying to give you an easy example with a function to understand why it's not true :) – user380364 Aug 30 '18 at 15:59
  • They are the same thing. And inside the context of a proof, you can only use finitely many. So by saying "for $i\in\Bbb N$, let $x_i$ be ...", you're saying that $\exists x_0\exists x_1\exists x_2\dots$. This will not end well for your proof. Yes, there are even more subtleties there, but understanding this part is a good start. – Asaf Karagila Aug 30 '18 at 16:14
  • Thanks a lot for your answers. I'll think about all this :) – user380364 Aug 30 '18 at 16:15
  • I think I can understand my error. Let $V={(x_1,x_2,...)\mid x_i\in\mathbb R }$. I defined the projection $p_i:V\to \mathbb R$ as $p_i(x_1,x_2,...)= x_i$. Let $(x_n)$ a sequence of $V$ and $x\to V$. I know that if $V$ has the product topology, then $x_n\to x \iff p_i(x_n)\to p_i(x)$. The implication $x_n\to x\implies p_i(x_n)\to p_i(x)$ is always true, but the converse may not be true as in Box topology for example (but it's true if $V$ has finite dimension). – user380364 Aug 30 '18 at 18:03
  • And I think it's a little bit my mistakes here in the sense that for finite collection of $A_i:= f^{-1}(i)$ it works, but I can't pick an element $(x_1,x_2,...)\in \prod_{i=1}^\infty f^{-1}(i)$ assuming that for all $m$ $\prod_{i=1}^m f^{-1}(i)\neq \emptyset$. Do you think the comparaison work ? And the axiom of choice say in fact that if I accept it, then my argument work. – user380364 Aug 30 '18 at 18:04
  • The axiom of choice, while very important in topology, is not really relevant to these kind of problems. In the finite case the product and box topologies coincide, for obvious reasons. In the infinite case, they do not. – Asaf Karagila Aug 30 '18 at 18:56
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When you chose an arbitrary preimage $a_b$ from the set $f^{-1}(b)$ of preimages for every $b\in B,$ you made an infinite number of arbitrary choices (assuming $B$ is infinite of course).

spaceisdarkgreen
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  • I don't get it (sorry)... this is the definition of surjectivity. I know that such $a_b$ exist for all $b$ by definition, so what's the matter ? – user380364 Aug 30 '18 at 13:17
  • Yes, you know the set of preimages is nonempty, but there may be many you have no canonical way of choosing a preimage. This is the essence of the axiom of choice. – spaceisdarkgreen Aug 30 '18 at 13:22
  • I'm so sorry, but I really have problem to understand. Moreover, what do you mean by "there may be many you have no canonical way of choosing a preimage." ? – user380364 Aug 30 '18 at 13:24
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    Imagine you have a family of nonempty sets $A_i$ for $i\in I.$ You say, for each $i,$ there is an $a_i\in A_i$ by definition, and thus there is a function that assigns each $A_i$ an element of it. This is the axiom of choice and exactly the same argument you made here. – spaceisdarkgreen Aug 30 '18 at 13:27
  • By ‘canonical’ I mean there is some well-defined way of specifying a unique element. Bertrand Russell gave a famous example: if you have an infinite number of pairs of shoes you don’t need to use the axiom of choice to get a shoe from each pair, whereas if you have an infinite number of pairs of socks, you do. If you think “the axiom of choice seems obviously true” from this perspective and are amazed it can imply very counterintuitive things, you are on your way to understanding. – spaceisdarkgreen Aug 30 '18 at 13:33
  • Thanks a lot for your answers, I really appreciate. I have to think a little bit more about all this :) It looks so weird for me to think that if ${A_i}_{i\in I}$ is a collection of non empty set, it could happen that $\forall i\in I, \exists a_i\in A_i$ can be not true (for me it's really the definition of "being a non empty set". In fact my version a choice axiom I have is a bit different, but it looks equivalent (I guess). – user380364 Aug 30 '18 at 13:37
  • This is the subtlety: $\forall i\in I,\exists a_i\in A_i$ is absolutely true by definition. What you need the axiom of choice for is to say there is a function $f$ from the collection of all $A_i$ to their union such that for all $i$, $f(A_i)\in A_i.$ Your impulse to say “just pick ‘the’ $a_i$ that exists” is precisely what invoking the axiom of choice means. There is no ‘the’ and unless you can find an explicit way to single one out (or there are only a finite number of $i$), you are using the axiom. – spaceisdarkgreen Aug 30 '18 at 14:12
  • If $\forall i\in I, \exists a_i\in A_i$ true, then let $f:{A_i}\to \bigcup_{i\in I}A_i$ defined by $$f(A_i)=a_i.$$ Here is your choice function, no ? Why it doesn't work ? – user380364 Aug 30 '18 at 16:14
  • @user380364: Suppose that $A_i$ is a set with two elements for each $i$, which of the two elements is $a_i$? – Asaf Karagila Aug 30 '18 at 16:24
  • @AsafKaragila: Thank you for answering. I would say "naively" why is this important ? Since such an $a_i$ exist, just take one... (I saw on your webpage that you are a specialist of the axiom of choice. I guess it's so obvious for you that such thing don't work... but in my point of view, it's so hard to imagine that such thing doesn't work...) May be you could give me an example so that I can understand better ? It's very frustrating to not understand. – user380364 Aug 30 '18 at 16:41
  • @user380364: Because otherwise you're just postulating that a choice function exists without justifying it. You claim that you've "constructed the choice function" by literally choosing an element from each set "separately" and claiming that this is done without choice. Yes, you choose one element without choice. But when you have to choose from infinitely many sets at once, you are effectively appealing to the axiom of choice. As for examples, it's hard to give instructive examples in the space of a comment. But try $F(r)={s\in\Bbb R\mid r-s\in\Bbb Q}$. What's the right inverse? – Asaf Karagila Aug 30 '18 at 16:45
  • @user380364 Your impulse that there is nothing wrong with your reasoning is simply the impulse most people have that the axiom of choice is obviously true. The axiom of choice is just a formalization of your reasoning here. If the axiom of choice is allowed, then there is nothing wrong with your reasoning, since your reasoning is the axiom of choice. For this reason, there is no example that “doesn’t work” though Asaf gives an example that gives you a non-measurable set, showing that it can lead to things that seem counterintuitive. – spaceisdarkgreen Aug 30 '18 at 17:55
  • @user380364 As for why it doesn’t follow from more basic considerations and axioms, your proofs must be finite so you cannot invoke existential instantiation an infinite number of times, which is the direction naive intuition would take you. Fraenkel/Mostowski/Cohen also showed it doesn’t follow from any more roundabout tour through other logical and set theoretical precepts. (At this point I’m just repeating what Asaf said.) – spaceisdarkgreen Aug 30 '18 at 18:07
  • @spaceisdarkgreen: thanks a lot for your explanations. I think that things are more clear now :) – user380364 Aug 30 '18 at 18:52