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    • CommentRowNumber1.
    • CommentAuthorzskoda
    • CommentTimeJun 26th 2011

    I added more details in essentially surjective functor. Please check for details (Mike?).

    • CommentRowNumber2.
    • CommentAuthorzskoda
    • CommentTimeJun 26th 2011

    New entry in my personal lab, somewhat related to the above: compatible idempotent monads (zoranskoda)

    • CommentRowNumber3.
    • CommentAuthorMike Shulman
    • CommentTimeJun 26th 2011

    Thanks! I reorganized, added links to extant pages, and moved the material about bijective-on-objects functors to bjective on objects functor and bo-ff factorization system. I also removed

    if an essentially surjective functor ff factorizes as jsj s where ss is surjective on objects then jj is an equivalence of categories.

    which doesn’t seem right at all to me, but I’m not sure what you meant to say.

    • CommentRowNumber4.
    • CommentAuthorzskoda
    • CommentTimeJun 27th 2011
    • (edited Jun 27th 2011)

    I meant

    if an essentially surjective functor ff factorizes as jsj s where ss is surjective on objects AND jj fully faithful, then jj is an equivalence of categories.

    Agreeing now ?

    By the way – typo in the link in your post above – “bjective”.

    • CommentRowNumber5.
    • CommentAuthorMike Shulman
    • CommentTimeJun 27th 2011

    Okay, now I agree that it’s true. (-: I wouldn’t be inclined to include it as a “property of essentially surjective functors”, though, since it decomposes into two more “basic” properties: (1) if f=jsf = j s and ff is essentially surjective, then so is jj, and (2) an essentially surjective and fully faithful functor is an equivalence. Note that surjectivity of ss is not needed.

    • CommentRowNumber6.
    • CommentAuthorzskoda
    • CommentTimeJun 27th 2011

    Well the intention is that b.o. is e.s.o. up to equivalence and that was just making more precise version of one of the directions. I am not sure what is the best thing to organize.

    • CommentRowNumber7.
    • CommentAuthorMike Shulman
    • CommentTimeJun 27th 2011

    We could say “any functor which is equivalent to a b.o. functor is e.s.o.”?

    • CommentRowNumber8.
    • CommentAuthorzskoda
    • CommentTimeJun 27th 2011
    • (edited Jun 27th 2011)

    Of course, but my intention was more, to treat both the strict factorization and weak factorization together and the difference is played precisely by insertint the mentioned equivalence. If b.o. and fully faithful make a strict factorization system then factorizing the same morphism as e.s.o. followed by the fully faithful is not unique as it amounts precisely to the replacement in the previous strict factorization system of the b.o. by any equivalent e.s.o. I agree that the surjectivity of ss is not needed above, but it is needed to make the factorization unique.

    If I am just complicating what in others’ thoughts is useless exercise of trivivialities, never mind.

    • CommentRowNumber9.
    • CommentAuthorMike Shulman
    • CommentTimeJun 28th 2011

    Side point: the factorization system involving eso functors is not a weak factorization system in the sense usually meant, rather a “2-categorical” or “up-to-isomorphism” factorization system.

    I added another paragraph to the page; does this at least partially say what you are getting at?

    • CommentRowNumber10.
    • CommentAuthorzskoda
    • CommentTimeJun 28th 2011

    Right, this looks great. Thanks also for correcting my incorrect wording “weak factorization system”.

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