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1. • CommentRowNumber1.
   • CommentAuthorYaron
   • CommentTimeDec 2nd 2010
   • PermaLink
   Author: Yaron
   Format: MarkdownItexAdded an entry on [[cocompleteness of varieties of algebras]]. It surely needs some improvements, but I hope that there are no fatal errors.

   Added an entry on cocompleteness of varieties of algebras. It surely needs some improvements, but I hope that there are no fatal errors.

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeDec 2nd 2010
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexYaron, I think this is fine, but there are alternative proofs which you might enjoy. First, the category of algebras of a monad $T$ on $Set$ has coequalizers; see the proof of proposition 3.4 (page 278 of 303) <a href="http://www.case.edu/artsci/math/wells/pub/pdf/ttt.pdf">here</a>. Then, you can get coproducts easily: let $U: Alg_T \to Set$ be the underlying functor, with left adjoint $F$. If you have a family of $T$-algebras $\{A_i\}$, then there are canonical coequalizers $$F U F U A_i \stackrel{\to}{\to} F U A_i \to A_i$$ and since $F(\sum_i U A_i)$ is the coproduct $\sum_i F U A_i$ in the category of algebras, i.e., since coproducts of free algebras exist, the coproduct of the $A_i$ is constructed as a coequalizer of an evident pair $$\sum_i F U F U A_i \stackrel{\to}{\to} \sum_i F U A_i$$ obtained by summing over all the canonical pairs.

   Yaron, I think this is fine, but there are alternative proofs which you might enjoy. First, the category of algebras of a monad $T$ on $Set$ has coequalizers; see the proof of proposition 3.4 (page 278 of 303) here. Then, you can get coproducts easily: let $U: Alg_T \to Set$ be the underlying functor, with left adjoint $F$. If you have a family of $T$-algebras $\{A_i\}$, then there are canonical coequalizers

   $$F U F U A_i \stackrel{\to}{\to} F U A_i \to A_i$$

   and since $F(\sum_i U A_i)$ is the coproduct $\sum_i F U A_i$ in the category of algebras, i.e., since coproducts of free algebras exist, the coproduct of the $A_i$ is constructed as a coequalizer of an evident pair

   $$\sum_i F U F U A_i \stackrel{\to}{\to} \sum_i F U A_i$$

   obtained by summing over all the canonical pairs.

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeDec 2nd 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItex> see the proof of proposition 3.4 (page 278 of 303) here and add it to the nLab!

   see the proof of proposition 3.4 (page 278 of 303) here

   and add it to the nLab!

4. • CommentRowNumber4.
   • CommentAuthorTodd_Trimble
   • CommentTimeDec 2nd 2010
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItex> and add it to the nLab! Done. I added two sections: (a), that the category of algebras of a monad on $Set$ is cocomplete, and (b), that the category of algebras of a finitary monad on a cocomplete cartesian closed category is cocomplete. These could be useful in practice.

   and add it to the nLab!

   Done. I added two sections: (a), that the category of algebras of a monad on $Set$ is cocomplete, and (b), that the category of algebras of a finitary monad on a cocomplete cartesian closed category is cocomplete. These could be useful in practice.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeDec 2nd 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItex> Done. Thanks! . Very nice. Let's try to make sure that this entry is being linked to from where it needs to be linked to, so that one can find it when looking for information on this question.

   Done.

   Thanks! . Very nice.

   Let’s try to make sure that this entry is being linked to from where it needs to be linked to, so that one can find it when looking for information on this question.

6. • CommentRowNumber6.
   • CommentAuthorYaron
   • CommentTimeDec 2nd 2010
   • PermaLink
   Author: Yaron
   Format: MarkdownItexTodd, thanks for the explanation and the references! It will take me some time to understand this (that is, why the coequalizer of the last pair is the required coproduct). Also, just to be sure, is it true that the assumption that the monad is on $\mathbf{Set}$ was used only for (1) getting the coequalizers for the monad by the Theorem of Barr-Wells, and (2) For getting the coproducts of the free algebras? I'm asking this because I'm a little confused by a similar proposition in Borceux (Vol. 2, Prop. 4.3.4, p. 200). This proposition asserts that, for a monad $\mathbb{T}$ on a cocomplete category $\mathcal{C}$, the category of algebras $\mathcal{C}^{\mathbb{T}}$ is cocomplete iff it has coequlizers. So, it seems that he is exactly in the situation of your proposition (he assumes coequlizers, and by cocompleteness of $\mathcal{C}$, he has coproducts of free objects), but his proof for the existence of the coproduct is more than 2 pages long. What's going on here? :)

   Todd, thanks for the explanation and the references! It will take me some time to understand this (that is, why the coequalizer of the last pair is the required coproduct).

   Also, just to be sure, is it true that the assumption that the monad is on $\mathbf{Set}$ was used only for (1) getting the coequalizers for the monad by the Theorem of Barr-Wells, and (2) For getting the coproducts of the free algebras?

   I’m asking this because I’m a little confused by a similar proposition in Borceux (Vol. 2, Prop. 4.3.4, p. 200). This proposition asserts that, for a monad $\mathbb{T}$ on a cocomplete category $\mathcal{C}$, the category of algebras $\mathcal{C}^{\mathbb{T}}$ is cocomplete iff it has coequlizers.

   So, it seems that he is exactly in the situation of your proposition (he assumes coequlizers, and by cocompleteness of $\mathcal{C}$, he has coproducts of free objects), but his proof for the existence of the coproduct is more than 2 pages long. What’s going on here? :)

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeDec 2nd 2010
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   Author: Mike Shulman
   Format: MarkdownItexAn original reference is "Coequalizers in categories of algebras" by Fred Linton (1969), in which he proves (essentially) that 1. If $C^T$ has reflexive coequalizers, then it has all colimits that C does (by the same argument Todd gave, which is also the same as Borceaux's—Borceaux just spells everything out explicitly, making a bit less use of universal properties); 1. If C has reflexive coequalizers and T preserves them, then $C^T$ has them (this is a special case of the general fact that $C^T$ inherits any colimits which C has and T preserves); 1. if C has an (epi,mono) factorization system preserved by T, has small products and is well-copowered, then $C^T$ has coequalizers; and 1. any monad on Set preserves (epi,mono) factorizations (essentially since assuming AC, all epis in Set are split, and almost all monos in Set are split). I would think the general facts about algebras for a monad would fit better at [[Eilenberg-Moore category]], maybe, or some other page not specifically titled as being about varieties of algebras.

   An original reference is “Coequalizers in categories of algebras” by Fred Linton (1969), in which he proves (essentially) that

   1. If $C^T$ has reflexive coequalizers, then it has all colimits that C does (by the same argument Todd gave, which is also the same as Borceaux’s—Borceaux just spells everything out explicitly, making a bit less use of universal properties);
   2. If C has reflexive coequalizers and T preserves them, then $C^T$ has them (this is a special case of the general fact that $C^T$ inherits any colimits which C has and T preserves);
   3. if C has an (epi,mono) factorization system preserved by T, has small products and is well-copowered, then $C^T$ has coequalizers; and
   4. any monad on Set preserves (epi,mono) factorizations (essentially since assuming AC, all epis in Set are split, and almost all monos in Set are split).

   I would think the general facts about algebras for a monad would fit better at Eilenberg-Moore category, maybe, or some other page not specifically titled as being about varieties of algebras.

8. • CommentRowNumber8.
   • CommentAuthorYaron
   • CommentTimeDec 3rd 2010
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   Author: Yaron
   Format: MarkdownItexMike, thanks for the reference to Linton and for the review of his results. In view of what you say, I think that as a first step I'll just read the proof in Borceux.

   Mike, thanks for the reference to Linton and for the review of his results. In view of what you say, I think that as a first step I’ll just read the proof in Borceux.

9. • CommentRowNumber9.
   • CommentAuthorTodd_Trimble
   • CommentTimeDec 3rd 2010
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexThanks, Yaron. Let me run through the bit about the coequalizer of the last pair being the coproduct. So, suppose $h: \sum_i F U A_i \to B$ is an algebra map which coequalizes that pair of arrows; this map corresponds to an $i$-indexed tuple of maps $h_i: F U A_i \to B$. Now consider the composite $$\sum_i F U F U A_i \stackrel{\sum_i F U \varepsilon A_i}{\to} \sum_i F U A_i \stackrel{h}{\to} B$$ The [[adjunct]] of this is the ($i$-indexed) family of maps $$F U F U A_i \stackrel{F U \varepsilon A_i}{\to} F U A_i \stackrel{h_i}{\to} B$$ By similar reasoning, the adjunct of the other composite is the family $$F U F U A_i \stackrel{\varepsilon F U A_i}{\to} F U A_i \stackrel{h_i}{\to} B$$ and these two adjuncts must be equal. So $h_i: F U A_i \to B$ factors through a unique algebra map $f_i: A_i \to B$ for all $i$, and we have thus shown that maps out of the coequalizer corresponds to families of algebra maps $f_i: A_i \to B$, making the coequalizer the coproduct.

   Thanks, Yaron. Let me run through the bit about the coequalizer of the last pair being the coproduct.

   So, suppose $h: \sum_i F U A_i \to B$ is an algebra map which coequalizes that pair of arrows; this map corresponds to an $i$-indexed tuple of maps $h_i: F U A_i \to B$. Now consider the composite

   $$\sum_i F U F U A_i \stackrel{\sum_i F U \varepsilon A_i}{\to} \sum_i F U A_i \stackrel{h}{\to} B$$

   The adjunct of this is the ($i$-indexed) family of maps

   $$F U F U A_i \stackrel{F U \varepsilon A_i}{\to} F U A_i \stackrel{h_i}{\to} B$$

   By similar reasoning, the adjunct of the other composite is the family

   $$F U F U A_i \stackrel{\varepsilon F U A_i}{\to} F U A_i \stackrel{h_i}{\to} B$$

   and these two adjuncts must be equal. So $h_i: F U A_i \to B$ factors through a unique algebra map $f_i: A_i \to B$ for all $i$, and we have thus shown that maps out of the coequalizer corresponds to families of algebra maps $f_i: A_i \to B$, making the coequalizer the coproduct.

10. • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeDec 3rd 2010
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    Author: Urs
    Format: MarkdownItexhere is the broken record again: please somebody make sure to give all these new bits of information their proper place on the lab. Thanks!

    here is the broken record again: please somebody make sure to give all these new bits of information their proper place on the lab. Thanks!

11. • CommentRowNumber11.
    • CommentAuthorYaron
    • CommentTimeDec 3rd 2010
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    Author: Yaron
    Format: MarkdownItexTodd, thanks for the detailed explanation! I've just finished reading the proof in Borceux, and I'll now read carefully your explanation.

    Todd, thanks for the detailed explanation! I’ve just finished reading the proof in Borceux, and I’ll now read carefully your explanation.

12. • CommentRowNumber12.
    • CommentAuthorTodd_Trimble
    • CommentTimeOct 3rd 2015
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexAdded some more material to [[colimits in categories of algebras]], leading up to the conclusion that categories monadic over $Set$ (or over $Set/X$, covering the multisorted case) are Barr-exact.

    Added some more material to colimits in categories of algebras, leading up to the conclusion that categories monadic over $Set$ (or over $Set/X$, covering the multisorted case) are Barr-exact.

13. • CommentRowNumber13.
    • CommentAuthorTodd_Trimble
    • CommentTimeOct 11th 2015
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexMore edits at [[colimits in categories of algebras]], trying to keep in mind results likely to be useful to *customers* of category theory. I think I'm done for the time being.

    More edits at colimits in categories of algebras, trying to keep in mind results likely to be useful to customers of category theory. I think I’m done for the time being.

14. • CommentRowNumber14.
    • CommentAuthorUrs
    • CommentTimeOct 12th 2015
    • PermaLink
    Author: Urs
    Format: MarkdownItexThanks, that's an beautiful page! I went there and turned many of the \to in the diagrams into \longrightarrow Because, at least on my system (but I think that's MathJax behaviour in general), the former doesn't render as intended in \stackrel stacking (it comes out too short).

    Thanks, that’s an beautiful page!

    I went there and turned many of the

     \to
    

    in the diagrams into

     \longrightarrow
    

    Because, at least on my system (but I think that’s MathJax behaviour in general), the former doesn’t render as intended in

     \stackrel
    

    stacking (it comes out too short).