How does the definition of a type of algebraic structures in Stacks Projects differ to the usual definition in universal algebra?

Question

the Stacks Project defines an algebraic structure (see tag 007L) as a category $\mathcal{C}$ together with a functor $F \colon \mathcal{C} \to \mathbf{Set}$ satisfying the following properties:

1. F is faithful.
2. $\mathcal{C}$ is complete and $F$ is continuous.
3. $\mathcal{C}$ admits filtered colimits and $F$ commutes with them.
4. $F$ reflects isomorphism.

How does this definition relates to the usual definition (see the end of the post) of the category of algebraic structures of a given type $\tau$ in universal algebra? To avoid confusion I will call them $\tau$-algebras from now on.

I know that the category of $\tau$-algebras for a given type $\tau$ satisfy all the properties above, but is the definition above equivalent? More precisly, given an algebraic structure $(\mathcal{C}, F)$ as above, is there a type $\tau$ such that the category of $\tau$-algebras is equivalent to $\mathcal{C}$ respecting $F$.

To avoid confusion I shortly recall the definition of a$\tau$-algebra I have in mind:

A type $(\Omega, E)$ consists of

• a set $\Omega$ (whose elements are called operators),
• a map $\nu \colon \Omega \to \mathbb{N}_0$ where the image of an operator is called the arity of it,
• and set $E$ of equations,

where an equation is an ordered pair of terms in $X_1, \ldots, X_m$ and a term in $X_1, \ldots, X_m$ is a symbolic finite pattern build by the following rules:

• $X_i$ for $1 \leq i \leq m$ is a term in $X_1, \ldots, X_m$.
• If $\omega \in \Omega$ and $T_1, \ldots, T_{\nu(\omega)}$ are terms in $X_1, \ldots, X_m$, then $\omega(T_1, \ldots, T_{\nu(\omega)})$ is also a term in $X_1, \ldots, X_m$.
• Those are the only possible ways to build a term in $X_1, \ldots, X_m$.

Now let $\tau = (\Omega, E)$ be a type. A $\tau$-algebra consists of a set $A$ and a family of maps $$ (f_\omega \colon A^{\nu(\omega)} \to A)_{\omega \in \Omega} $$ satisfying the equations, i.e.:

Let $T$ be a term $X_1, \ldots, X_m$. Then for all $a \in A^m$ we can assign a value $|T|(x) \in X$ to $T$ recursively defined by

• $|X_i|(a) = a_i$ and
• $|\omega(T_1, \ldots, T_n)|(a) = f_\omega(|T_1|(a), \ldots, |T_n|(a))$

Let $(T,S)$ be an equation where $S,T$ are terms in $X_1, \ldots, X_m$ then we require $$ |T|(a) = |S|(a) $$ for all $a \in A^m$.

Answer 1

I'll rename $F$ to $U$ (for "underlying"). This definition is equivalent to the universal algebra definition if we require either that $C$ is locally presentable or that $U$ has a left adjoint (call this axiom 2'), and also replace filtered colimits in axiom 3 with sifted colimits (call this axiom 3'), as follows:

1. If $C$ is locally presentable then $C$ has limits, $U$ preserves them, and $U$ is accessible, so by the locally presentable adjoint functor theorem, $U$ has a left adjoint $F : \text{Set} \to C$ ($F$ for "free"). Alternatively we can just require that $U$ has a left adjoint in the axiomatization.

2. By the crude monadicity theorem, $U$ has a left adjoint, is conservative, and preserves reflexive coequalizers (using the fact that reflexive coequalizers are sifted), hence the adjunction $F \vdash U$ is monadic. So $C$ is the category of algebras of a monad $UF$ on $\text{Set}$.

3. Since $U$ preserves filtered colimits and $F$ is cocontinuous, $UF$ preserves filtered colimits and is therefore a finitary monad. Finitary monads on $\text{Set}$ are in turn equivalent to Lawvere theories, and categories of models of Lawvere theories are exactly universal algebraic structures in the sense you describe. Moreover finitary monads on $\text{Set}$ preserve sifted colimits and their categories of algebras are locally presentable, so this definition is equivalent to the modification with 2' and 3'.

Every example the Stacks project lists in tag 007L, and I bet every example they discuss anywhere, is the category of models of a Lawvere theory and so satisfies the stronger axioms 2' and 3'. Personally I find "category of models of a Lawvere theory" to be the cleanest way of describing algebraic structures and I don't understand why the Stacks project is using such a bizarre axiomatization. With the Stacks project's axiomatization you a priori need to verify all of these axioms by hand for each example, but instead you can just prove once and for all that

1. categories of models of Lawvere theories satisfy all of these axioms, and

2. categories of algebraic structures in the sense of universal algebra are categories of models of Lawvere theories.

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This argument implies that the Stacks project's definition could be strictly more general than universal algebra in two different ways: $U$ could fail to have a left adjoint (which implies that $C$ is not locally presentable), or $U$ could have a left adjoint but the adjunction $F \vdash U$ could fail to be monadic (which implies that $U$ does not preserve reflexive coequalizers and so does not preserve sifted colimits).

An example of the first situation is the following: take $C$ to be the category of $M$-sets for $M$ a "large monoid" (a "monoid" whose underlying "set" is a proper class, say the free monoid on a proper class to be specific). Limits and colimits are computed as in $\text{Set}$, so the forgetful functor $U : C \to \text{Set}$ preserves them, and $U$ is also faithful and conservative. But $U$ does not have a left adjoint because $M$ is not a set, and $C$ is not locally presentable for the same reason.

I don't know an example of the second situation off the top of my head. It seems tricky to write down an example where $U$ preserves filtered colimits but not reflexive coequalizers. Maybe a finite limit sketch could be used here.

Edit: I think $C = \text{Cat}$ (the category of small categories) and $U : \text{Cat} \to \text{Set}$ the forgetful functor given by sending a small category to the disjoint union of its set of objects and its set of morphisms is an example but I am too lazy to check explicitly that $U$ preserves filtered colimits but not reflexive coequalizers; anyone else want to take a stab?