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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 $i$ is a section for $p$, I understand that $p$ is an absolute coequalizer of $ip, 1: UB \overset{\to}{\to} UB$, so that $Tp$ is a coequalizer for $Tip, 1, TUB \overset{\to}{\to} TUB$. But I don’t understand how this implies that $Tp$ is a coequalizer for $TU\pi_1, TU\pi_2: TUE \overset{\to}{\to} TUB$.

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 $\mathsf{Set}$ (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 $Set$ 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. $p$ is a coequalizer for $(U\pi_1,U\pi_2)$.
2. Take a section $i$ for $p$. Then $p$ is also an absolute coequalizer for $(ip,1)$.
3. So the monadic functor $U: T-Alg \to Set$ creates this absolute coequalizer in $T-Alg$.
4. If $f: B \to X$ coequalizes $(U\pi_1,U\pi_2)$, then $f$ factors uniquely as $f = \bar f p$.
5. It follows that $f$ coequalizes $(ip,1)$. So $\bar f$ 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 $Set^T$ :-), then the crucial thing which might need spelling out in that article is that if $\langle \pi_1, \pi_2 \rangle: E \hookrightarrow X \times X$ is an equivalence relation in $Set$ and

$E \stackrel{\overset{\pi_1}{\longrightarrow}}{\underset{\pi_2}{\longrightarrow}} X \stackrel{q}{\to} Y$

is exact, then this fork splits. Proof: choose a section $s$ of $q$. This gives a basepoint $s(y)$ for each equivalence class $y \in Y = X/E$. A splitting of the fork is given by $s$ together with $X \to E: x \mapsto (x, s q(x)) \in E$. Of course, split forks are absolute coequalizers, so $T$ 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 $Set$ 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 $Set$.

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 $\langle ip, 1 \rangle : B \to E$ could really denote a map into $E$ rather than $B \times B$, so I misinterpreted it as meaning a pair of maps $B \to B$ 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 $UB \to UE$ instead of $B \to E$ 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 $C$ has (reflexive) coequalizers whose quotient maps split, then $C^T$ has (reflexive) coequalizers for any monad $T$. 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.