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?
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ I don't know a published reference, but I gave a proof at math.stackexchange.com/questions/1558623/…. $\endgroup$– Eric WofseyNov 29, 2017 at 4:57
-
1$\begingroup$ You should pick a more descriptive title. $\endgroup$– Asaf Karagila ♦Nov 29, 2017 at 5:01
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
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
answered Nov 29, 2017 at 4:59
-
$\begingroup$ 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
-
1$\begingroup$ And there is a lot more to this paper than the "rather trivial result" enquired about here. $\endgroup$ Nov 29, 2017 at 5:16
$\begingroup$
$\endgroup$
3
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.
-
$\begingroup$ You probably want your sets to be pairwise disjoint. $\endgroup$– Asaf Karagila ♦Nov 29, 2017 at 6:06
-
-