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    • CommentRowNumber1.
    • CommentAuthorMike Shulman
    • CommentTimeAug 18th 2013

    I’m not entirely happy with the introduction (“Statement”) to the page axiom of choice. On the one hand, it implies that the axiom of choice is something to be considered relative to a given category CC (which is reasonable), but it then proceeds to give the external formulation of AC for such a CC, which I think is usually not the best meaning of “AC relative to CC”. I would prefer to give the Statement as “every surjection in the category of sets splits” and then discuss later that analogous statements for other categories (including both internal and external ones) can also be called “axioms of choice” — but with emphasis on the internal ones, since they are what correspond to the original axiom of choice (for sets) in the internal logic.

    (I would also prefer to change “epimorphism” for “surjection” or “regular/effective epimorphism”, especially when generalizing away from sets.)

    • CommentRowNumber2.
    • CommentAuthorTobyBartels
    • CommentTimeAug 19th 2013

    I agree.

    • CommentRowNumber3.
    • CommentAuthorMike Shulman
    • CommentTimeAug 24th 2013

    Since no one objected, I went ahead and made this change.

    • CommentRowNumber4.
    • CommentAuthorThomas Holder
    • CommentTimeOct 16th 2018

    I added the characterization of IAC toposes as Boolean étendues.

    diff, v66, current

    • CommentRowNumber5.
    • CommentAuthorMike Shulman
    • CommentTimeMay 2nd 2020

    Added the fact (thanks to Alizter for finding it) that AC is equivalent to the statement “if two free groups have equal cardinality, then so do their generating sets”.

    diff, v70, current

    • CommentRowNumber6.
    • CommentAuthorMike Shulman
    • CommentTimeMay 2nd 2020

    Oops, of course the sets have to be infinite.

    diff, v70, current

    • CommentRowNumber7.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 2nd 2020

    Seems like a somewhat roundabout way of putting it: can’t we just say that for infinite XX, that XX and F(X)F(X) have the same cardinality? Am I missing something?

    • CommentRowNumber8.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 2nd 2020

    Okay, I guess never mind my question. The direction that the indicated statement plus ZF implies AC doesn’t look easy.

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