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Let $\mathrm{AlgTh}$ be the category of one-sorted algebraic theories (synonym: Lawvere theories; morphisms are functors that are identical on objects and strictly preserve products). It is known that it is locally representable, hence it is bicomplete.

I wonder how the limits and colimits are described in it.

I know that (see e.g. The Category Theoretic Understanding of Universal Algebra: Lawvere Theories and Monads)

  1. The initial object is an empty theory (algebras over it are sets)
  2. The terminal object is a trivial theory (the category of algebras is $1$)
  3. The coproduct of two theories is a disjoint union of the operations of theories with the preservation of internal axioms, without adding new ones

How are products, equalizers, and coequalizers described?

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Typically, limits of multisorted algebraic theories (by which I mean a pair of a set $S$ and an $S$-sorted algebraic theory $\mathbb F(S) \to L$) are most easily described in terms of their presentation as categories with finite products (i.e. a cartesian (monoidal) category), whereas colimits are most easily described in terms of their presentation as equational presentations, i.e. by sets of operations and equations.

  • The limit of a diagram of multisorted algebraic theories is given by the limit of the underlying cartesian categories. This means, for instance, that the product of an $S$-sorted algebraic theory with an $S'$-sorted algebraic theory will be an $(S \times S')$-sorted algebraic theory. When $S = S' = 1$, this recovers the product of one-sorted algebraic theories, since $1 \times 1 \cong 1$. Limits of algebraic theories typically do not have nice descriptions in terms of the operations and equations (e.g. the product of two finitely presented algebraic theories will not usually be finitely presented).
  • Take a morphism of multisorted algebraic theories, from an $S$-sorted algebraic theory $\mathbb F(S) \to L$ to an $S'$-sorted algebraic theory $\mathbb F(S') \to L'$, to comprise a pair $(s, f)$ of a function $s : S \to S'$ and a functor between the codomains $f : L \to L'$ forming an evident commutative square*. Syntactically $f : L \to L'$ maps $S$-sorted terms of $L$ to $S'$-sorted terms of $L'$. The coequaliser of two such morphisms is given by the $S'$-sorted algebraic theory $L''$ obtained by imposing equations on $L'$ that equate any two terms whose preimage under $f$ is equal. When $S = S' = 1$, this recovers the coequaliser of one-sorted algebraic theories.

(*More generally, we could take $s$ to be a function $S \to |\mathbb F(S')|$, but this definition is a little less intuitive syntactically.)

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  • $\begingroup$ Thank you! Just don't delete it please! The part about multisorts is also interesting! And what morphisms of multisorted theories do you mean? $\endgroup$ Jun 9, 2023 at 10:55
  • $\begingroup$ I've updated my answer. Hopefully this clarifies the situation. $\endgroup$
    – varkor
    Jun 9, 2023 at 11:11
  • $\begingroup$ Thank you very much! Is a cartesian category a category with finite limits? Do you mean that it is necessary to take the limit in the category of small cartesian categories? $\endgroup$ Jun 9, 2023 at 12:35
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    $\begingroup$ By "cartesian category", I mean a category with finite products (I've clarified this in my answer). Yes, you take the limit in the category of small categories with finite products (which is equivalently the limit in the category of small categories). $\endgroup$
    – varkor
    Jun 9, 2023 at 12:47

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