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Let $A$ and $B$ be sets. If there exists a surjection $f : A \to B$ then there exists an injection $g : B \to A$.

Proof: given $b \in B$ select an element $a \in f^{-1}(b)$. Denote this element by $g(b)$. Then $g(b) \in f^{-1}(b)$ so that $f(g(b)) = b$. Consequently $g(b_1) = g(b_2)$ implies $f(g(b_1)) = f(g(b_2))$ so that $b_1 = b_2$. We conclude $g$ is an injection. QED

Is the axiom of choice required to make this argument rigorous?

Umberto P.
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I don't have a proof that the existence of any injection depends on the axiom of choice, but the existence of a right inverse is equivalent to AC, as follows. Let $X$ be a set not containing $\emptyset$ and consider the set $Y = \{(z, A)\ |\ z\in A \in X\}$. Define $f:Y\rightarrow X$ by $f((z,A)) = A$. If $g$ is a right inverse of $f$, then $g^*(A)$ given by the the first element of the ordered pair $g(A)$ is a choice function.

universalset
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It is. In fact, I believe it can be show that the universal truth of this theorem is equivalent to the axiom of choice.

ncmathsadist
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  • In fact, this is equivalent to the axiom of choice. – ncmathsadist Dec 13 '13 at 01:45
  • It's clear that AC is equivalent to the statement "every surjection splits". But if the injection $g$ in the question doesn't split $f$, how do you construct the choice function? – Alex Kruckman Dec 13 '13 at 01:47
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    This is a nontrivial affair. See this: http://www.math.uwo.ca/~srankin/courses/4123/2011/zorns_lemma.pdf – ncmathsadist Dec 13 '13 at 01:50
  • Now I'm really confused... what does this have to do with Zorn's Lemma? – Alex Kruckman Dec 13 '13 at 01:55
  • Zorn's lemma is another equivalent of the axiom of choice. This construction is nontrivial. – ncmathsadist Dec 13 '13 at 02:14
  • http://planetmath.org/surjectionandaxiomofchoice – ncmathsadist Dec 13 '13 at 02:21
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    @ncmathsadist I'm well aware that Zorn's Lemma is equivalent to the axiom of choice, but I don't see how that's relevant to this discussion. Your link to planetmath just reiterates the fact that "every surjection splits" is equivalent to the axiom of choice, which I commented on above, and which is the content of the accepted answer. In Asaf's answer here http://math.stackexchange.com/a/192486/7062, he says that it's still open whether "surjection $X\rightarrow Y$ implies injection $Y\rightarrow X$" implies the axiom of choice, which is what you claim. – Alex Kruckman Dec 13 '13 at 02:44
  • I didn't say this was a trivial implication. – ncmathsadist Dec 13 '13 at 02:59
  • I would love to see your proof! – Alex Kruckman Dec 13 '13 at 03:41
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    I don't understand, are you saying that you have a proof that the partition principle implies the axiom if choice?? All the linked proofs do not contain such proof. – Asaf Karagila Dec 13 '13 at 10:14
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    By the way, the problem whether or not the partition principle implies the axiom of choice is the oldest open problem of set theory. I am not aware of any progress made towards it in the last twenty years. So if you have a proof, I would love to see that. – Asaf Karagila Dec 13 '13 at 10:21