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1$\begingroup$ Possibly related: mathoverflow.net/questions/279558/… mathoverflow.net/questions/273551/… $\endgroup$Alec Rhea– Alec Rhea2017-10-27 10:35:00 +00:00Commented Oct 27, 2017 at 10:35
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7$\begingroup$ Are you familiar with Makkai's paper "Avoiding the axiom of choice in general category theory" ? It is a bit orthogonal to what you mention in your question (As far as I remember it does not discuss universe) but it develop very good notions which avoid all those anoying problems of having to chose limits and so one... $\endgroup$Simon Henry– Simon Henry2017-10-27 10:35:58 +00:00Commented Oct 27, 2017 at 10:35
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3$\begingroup$ Yes, but I'm saying that the focus on universes is a bit irrelevant. It's just a small observation. There are other things which require the use of choice, in an actual and substantial way. Universes are entirely boring. Yes, you need to formulate exactly what you mean by an inaccessible... hurray... what else is new in the choiceless world? $\endgroup$Asaf Karagila– Asaf Karagila ♦2017-10-27 12:20:34 +00:00Commented Oct 27, 2017 at 12:20
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2$\begingroup$ @OmerRosler : roughly yes, more precesely an anafunctor is something to which object $c \in C$ attach a 'set' (even a class in some cases) of objects $F(c) \rightarrow D$ with canonical isomorphism between each of these objects (and the action on morphims as well). for example if you want to define the anafunctor $C \times C \rightarrow C$ which to each pairs of objects $C$ associate their products then you just associate the set of all possible products and their canonical isomorphisms between them instead of having to 'chose' one product for each pair. The theory works very nicely... $\endgroup$Simon Henry– Simon Henry2017-10-27 13:05:16 +00:00Commented Oct 27, 2017 at 13:05
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2$\begingroup$ ... The only place where you are getting into trouble is that without some very weak choice principles (discussed at the end of the paper. you end up with the fact that the category of anafunctors between two small category may fails to be small it self (see mathoverflow.net/questions/264585/… ) $\endgroup$Simon Henry– Simon Henry2017-10-27 13:07:37 +00:00Commented Oct 27, 2017 at 13:07
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