While trying to prove that the monad associated to a Lawvere theory is finitary, I came across the following, which troubles me.

Let $\mathcal A$ be small category and $\mathcal C$ be the full subcategory of $\widehat{\mathcal A}$ whose objects are those presheaves $\mathcal A^{\rm op} \to \mathsf{Set}$ which preserves finite products. Denote $i \colon \mathcal C \to \widehat{\mathcal A}$ the inclusion.

Proposition (probably false). The functor $i$ creates sifted colimits.

Proof. Take $D \colon \mathcal J \to \mathcal C$ a diagram over a sifted index category. Denote $M$ the colimit of $iD$ in $\widehat{\mathcal A}$. Then $M$ lies in $\mathcal C$: $$ \begin{aligned} M \left( \prod_{k=0}^n a_k \right) &\stackrel{(1)}\simeq \operatorname{colim}_{j\in J}\left( D(j)\left( \prod_{k=0}^n a_k \right) \right) \\ &\stackrel{(2)}\simeq \operatorname{colim}_{j\in J}\left( \prod_{k=0}^n D(j)\left( a_k \right) \right) \\ &\stackrel{(3)}\simeq \prod_{k=0}^n \operatorname{colim}_{j\in J}\left( D(j)\left( a_k \right) \right) \\ &\stackrel{(4)}\simeq \prod_{k=0}^n M\left( a_k \right) \end{aligned} $$ (1) because colimits in a presheaf category are computed point wise, (2) because each $D(j)$ is an object of $\mathcal C$, (3) because sifted colimits in $\mathsf{Set}$ commute with finite products, and (4) for the same reason than (1). $$ \hspace 330pt \square $$

But this can't be right, otherwise any monad associated to a Lawvere theory commutes with sifted colimits. As Lawvere theories are (essentially) in one to one correspondence with finitary monads, it seems very unlikely!

So, where is the flaw ?

  • 2
    $\begingroup$ As far as I know there is no flaw. Finitary monads preserve filtered colimits and reflexive coequalizers, and I believe this implies that they preserve sifted colimits. $\endgroup$ – Qiaochu Yuan Aug 10 '15 at 15:50
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    $\begingroup$ I'm not so certain that filtered colimits + reflexive coequalisers suffice for all sifted colimits, but at any rate the conclusion about finitary monads on $\mathbf{Set}$ is correct. It's less clear to me what happens for monads of higher accessibility rank, though. $\endgroup$ – Zhen Lin Aug 10 '15 at 16:08
  • $\begingroup$ @QiaochuYuan Ok, so finitary monads coincide with sifted-colimit-preserving monads (on $\mathsf{Set}$)? I can't say I have suspected that... But it is for the best, all trouble is now gone! If you or Zhen Lin want to convert that as an answer, I'll be happy to accept it. $\endgroup$ – Pece Aug 10 '15 at 19:23

In fact every (retract of) finitely generated free algebra $Fn$ is perfectly presentable, i.e. the hom-functor $\mathcal{A}(Fn, -)$ preserves sifted colimits. In particular, the forgetful functor $U = \mathcal{A}(F1, -)$ preserves sifted colimits so does the monad $UF$.

Also, a functor $F\colon \mathcal{A}\to\mathcal{B}$ preserves sifted colimit if and only if it is finitary and preserves reflexive coequalizers, whenever $\mathcal{A}$ is finitely cocomplete.

You may find details in

[1] Adámek, Rosický, Vitale. What are sifted colimits, TAC, Vol.23, No. 10, 2010

[2] Adámek, Rosický, Vitale. Algebraic Theories, 2010

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