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    • CommentRowNumber1.
    • CommentAuthorYaron
    • CommentTimeJan 6th 2011

    Todd, thanks for this interesting entry (colimits in categories of algebras)! I left a query box describing a possible simple proof for the last theorem using the adjoint lifting theorem (I hope there are no substantial errors).

    • CommentRowNumber2.
    • CommentAuthorTodd_Trimble
    • CommentTimeJan 6th 2011

    Yaron, yes you’re right (and thanks). If you’d like to go in and add this in as the official first proof, I’d be delighted. (But please don’t erase the pedestrian proof I wrote. I’ll rework that so that it looks like we’re giving a very concrete and hands-on description of the left adjoint; I think readers should be aware that this is not just some abstract black box or adjoint functor nonsense, but a construction they’re already familiar with from ordinary commutative algebra and module theory.)

    • CommentRowNumber3.
    • CommentAuthorYaron
    • CommentTimeJan 6th 2011

    Thanks for the answer, Todd. I’ll happily do the change (though it will take some time, since I have to go now).

    And of course I will not delete the existing proof. Actually, I was so eager to make it clear that the short proof is just an addition, that I re-edited the page twice only to add the word “also” (it can also be proved by…)
    :)

    • CommentRowNumber4.
    • CommentAuthorYaron
    • CommentTimeJan 6th 2011

    Done.

    • CommentRowNumber5.
    • CommentAuthorTim Campion
    • CommentTimeJul 10th 2016

    I’m having difficulty understanding Lemma 1. If ii is a section for pp, I understand that pp is an absolute coequalizer of ip,1:UBUBip, 1: UB \overset{\to}{\to} UB, so that TpTp is a coequalizer for Tip,1,TUBTUBTip, 1, TUB \overset{\to}{\to} TUB. But I don’t understand how this implies that TpTp is a coequalizer for TUπ 1,TUπ 2:TUETUBTU\pi_1, TU\pi_2: TUE \overset{\to}{\to} TUB.

    Somewhat relatedly, I fixed a typo in Corollary 1.

    I find Corollary 1 to be surprising because there is a characterization of categories which are monadic over Set\mathsf{Set} (Thm 4.4.5 in the Handbook of Categorical Algebra vol. 2, also discussed by Vitale), and I don’t see anything in the characterization that obviously implies the existence of all coequalizers (although some coproducts are implied).

    • CommentRowNumber6.
    • CommentAuthorTodd_Trimble
    • CommentTimeJul 10th 2016

    I meant a coequalizer of a split fork. Sorry.

    I am fairly sure that categories that are monadic over SetSet are cocomplete. This is probably in Manes’ book, and the result might be due to Linton.

    • CommentRowNumber7.
    • CommentAuthorMike Shulman
    • CommentTimeJul 10th 2016

    In fact, I believe that categories monadic over Set are not just cocomplete but total.

    • CommentRowNumber8.
    • CommentAuthorTodd_Trimble
    • CommentTimeJul 10th 2016

    Indeed.

    • CommentRowNumber9.
    • CommentAuthorTim Campion
    • CommentTimeJul 10th 2016

    Okay, I think I understand now.

    1. pp is a coequalizer for (Uπ 1,Uπ 2)(U\pi_1,U\pi_2).
    2. Take a section ii for pp. Then pp is also an absolute coequalizer for (ip,1)(ip,1).
    3. So the monadic functor U:TAlgSetU: T-Alg \to Set creates this absolute coequalizer in TAlgT-Alg.
    4. If f:BXf: B \to X coequalizes (Uπ 1,Uπ 2)(U\pi_1,U\pi_2), then ff factors uniquely as f=f¯pf = \bar f p.
    5. It follows that ff coequalizes (ip,1)(ip,1). So f¯\bar f is an algebra homomorphism.

    The fact I wasn’t aware of is that if a functor creates one coequalizer, then it creates any other coequalizer with the same quotient. It’s weird to me, partly because I’m not sure how to generalize this fact to other indexing shapes.

    • CommentRowNumber10.
    • CommentAuthorTodd_Trimble
    • CommentTimeJul 10th 2016
    • (edited Jul 11th 2016)

    If you’re willing to settle for mere cocompleteness of Set TSet^T :-), then the crucial thing which might need spelling out in that article is that if π 1,π 2:EX×X\langle \pi_1, \pi_2 \rangle: E \hookrightarrow X \times X is an equivalence relation in SetSet and

    Eπ 2π 1XqYE \stackrel{\overset{\pi_1}{\longrightarrow}}{\underset{\pi_2}{\longrightarrow}} X \stackrel{q}{\to} Y

    is exact, then this fork splits. Proof: choose a section ss of qq. This gives a basepoint s(y)s(y) for each equivalence class yY=X/Ey \in Y = X/E. A splitting of the fork is given by ss together with XE:x(x,sq(x))EX \to E: x \mapsto (x, s q(x)) \in E. Of course, split forks are absolute coequalizers, so TT will preserve them.

    Clear now?

    • CommentRowNumber11.
    • CommentAuthorTim Campion
    • CommentTimeJul 10th 2016

    I thought I posted this a little while ago, but apparently it didn’t go up:

    It’s clear to me now, thanks. I’m surprised to learn that any equivalence relation in SetSet is already split, and hence absolute. I guess I should be familiar with this fact, since it explains why quotients by congruence relations in monadic categories are computed as in SetSet.

    Looking back, that’s exactly what it the article already said. I think what threw me was the part I “fixed”: it didn’t occur to me that the bracket notation in ip,1:BE\langle ip, 1 \rangle : B \to E could really denote a map into EE rather than B×BB \times B, so I misinterpreted it as meaning a pair of maps BBB \to B forming the “fork” part of a split fork. I’m not sure it could really be any clearer, though. The one change I would make would be to write UBUEUB \to UE instead of BEB \to E since these maps live downstairs.

    The funny thing is that the proof I gave doesn’t require kernel pairs to exist: it in fact shows that if CC has (reflexive) coequalizers whose quotient maps split, then C TC^T has (reflexive) coequalizers for any monad TT. I’m not sure how this slight increase in generality could be useful, though.

    • CommentRowNumber12.
    • CommentAuthorTim Campion
    • CommentTimeJul 10th 2016

    Never mind, my own proof doesn’t make sense.

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