Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a problem from Tao's Analysis I.

We are asked to show that the axiom of choice is equivalent to the statement that for any sets $A$ and $B$ for which a surjection $g:B\to A$ exists, an injection $f:A\to B$ exists.

I have proven that the axiom of choice implies the statement , but I am having difficulty with the other implication. The hint given in the book is to use the previous exercise, which says that the axiom of choice is equivalent to the statement that if $I$ is a set, $\forall\alpha\in I$ $X_\alpha$ is a non-empty-set and $\forall \alpha,\beta\in I, \alpha\neq \beta\implies X_\alpha\cap X_\beta=\emptyset$, then there exists a set $Y$ with $\forall \alpha\in I ,|Y\cap X_\alpha|=1$.

To use the previous exercise we need disjoint sets, so we consider the sets $\{\alpha\}\times X_\alpha$ instead of just $X_\alpha$. I think then we should find a set $Y$ with $\forall \alpha \in I,|Y\cap(\{\alpha\}\times X_\alpha)|=1$, perhaps by using that there is some surjection between two sets and so by assumption some injection which goes the other way, whose image is suitable as a set $Y$. I think that is the idea, but I can't figure out exactly what these functions should be.

If I am incorrect, please correct me. Any help is appreciated.

share|cite|improve this question
Michael One, two, @Asaf Karagila is coming for you. – Git Gud Mar 27 '13 at 20:46
up vote 3 down vote accepted

The problem's statement would be correct if $f$ were required to be a right inverse of $g$, i.e., for each $a\in A$, $f(a)$ selects an element from the fiber $g^{-1}(\{a\})$. Without this additional requirement, the statement is not known to be equivalent to the axiom of choice, nor is it known to be strictly weaker. This open problem is sometimes called the "map on map in" problem.

share|cite|improve this answer
Alright, I see. With this additional condition on the injection I can solve it. When I looked up "map on map in" I couldn't find anything relevant, but I'm quite interested (although it's probably way over my head.) Could you point me to somewhere I could read about it? – user50407 Mar 27 '13 at 21:23
@Michael: I never really heard the "map on map in" terminology, but "The Partition Principle" is the assertion that whenever there is a surjection $f\colon A\to B$ there is an injection $g\colon B\to A$. As Andreas said, it is still open whether or not the partition principle implies the axiom of choice. – Asaf Karagila Mar 27 '13 at 21:32
@AsafKaragila Thanks, with your suggestion I have found lots to read about the Partition Principle :) – user50407 Mar 27 '13 at 21:38
@Michael: No problem. PP is one of the sources of ideas for me, it's a never-ending fountain of ideas. One of my aims is to prove it does not imply the axiom of choice (even though all evidence point otherwise). I am fairly certain this is going to remain as a dream, but I have a lot of work in between which should be enough to keep me busy for the next few years! :-) – Asaf Karagila Mar 27 '13 at 22:11

Let $\{X_\alpha:\alpha\in I\}$ be a family of pairwise disjoint non-empty sets indexed by a set $I$. Let $X=\bigcup_{\alpha\in I}X_\alpha$, and define $g:X\to I$ by setting $g(x)=\alpha$ iff $x\in X_\alpha$. Then $g$ is a surjection, so by hypothesis there is an injection $f:I\to X$ such that $f(\alpha)\in X_\alpha$ for each $\alpha\in I$. (Note: I’m assuming that you inadvertently omitted that last bit in stating the problem. If not, the following argument doesn’t work, and I’m not sure that the result is even true.) Let $Y=f[I]$; then $|Y\cap X_\alpha|=1$ for each $\alpha\in I$. By the previous exercise, the axiom of choice holds.

share|cite|improve this answer
Thanks for the answer. The part that I omitted from the problem was not actually part of the question in the book. Using it, I should be able to solve the problem. I need to think about whose answer I should accept, considering that both you and Andreas Blass posted at the same time and gave me the answers that I needed to finish the problem. – user50407 Mar 27 '13 at 21:33
@Michael: You’re welcome. – Brian M. Scott Mar 27 '13 at 21:40

HINT: If $\{A_i\mid i\in I\}$ is a family of disjoint sets, then $a\mapsto i$ such that $a\in A_i$ is a surjection.

share|cite|improve this answer
This was also the surjection I was thinking about. I discovered that my confusion was caused by the fact that the problem does not explicitly state that the injection must be the right inverse of the surjection. I think with that I should be able to solve it. Thanks anyway. – user50407 Mar 27 '13 at 21:14
You're welcome. Also, with that vote I got the Legendary badge, so thanks! – Asaf Karagila Mar 27 '13 at 21:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.