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I've read it on Wikipedia that in the category of Sets, the statement "every set is a projective object" is equivalent to the Axiom of Choice.

I'd like some references with this proof but I'm having trouble finding them. Could anyone help me?

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This is due to Andreas Blass,

Blass, Andreas "Injectivity, Projectivity, and the Axiom of Choice." Transactions of the American Mathematical Society, Vol. 255, (Nov., 1979), pp. 31-59

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  • $\begingroup$ Well, this result is mentioned in passing on the second page, but is that really the original source of it? It's a rather trivial result, that I imagine was known from the first moment anyone pondered what projective objects are in general categories. $\endgroup$ Nov 29, 2017 at 5:01
  • $\begingroup$ Well. That's the reference I know. Maybe Andreas will stop by and clarify. History is full of nearly trivial observations that were only made quite late. $\endgroup$
    – Asaf Karagila
    Nov 29, 2017 at 5:03
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    $\begingroup$ And there is a lot more to this paper than the "rather trivial result" enquired about here. $\endgroup$ Nov 29, 2017 at 5:16
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Let $E_{\alpha}$ be a collection of nonempty, pairwise disjoint sets, let $\Lambda$ be the index set, let $\cup_{\alpha \in \Lambda} E_{\alpha}$ be denoted by $E$. Consider the diagram with maps $id: \Lambda \rightarrow \Lambda$ and $e: E: \rightarrow \Lambda$ where $e(x)=\alpha \mbox{ if } x \in E_{\alpha}$, note that this function is well defined as the sets $E_{\alpha}$ are pairwise disjoint. Now $\Lambda$ projective implies there is a map $c : \Lambda \rightarrow E$ such that the diagram commutes which is essentially the axiom of choice. And Axiom of choice will imply there is such a map. Thus finishing the proof.

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  • $\begingroup$ You probably want your sets to be pairwise disjoint. $\endgroup$
    – Asaf Karagila
    Nov 29, 2017 at 6:06
  • $\begingroup$ Right forgot to put emphasis on that. $\endgroup$ Nov 29, 2017 at 6:09
  • $\begingroup$ Corrected, thanks. $\endgroup$ Nov 29, 2017 at 6:11

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