Lawvere theory

Remarks

In $\aleph_0^{\mathrm{op}}$, the object $n$ is the product of $n$ copies of $1$, with projections given by the inclusions $\set{1} \to \set{1, \ldots, n}$. So in $\mathcal{T}$ every object is a power $1^n$ of the object $1$. A model $M$ is determined by the set $M(1)$ and the maps $M(1)^n \to M(1)$ given by the morphisms $n \to 1$.

Example: for a variety $\mathcal{V}$ with free algebras $F(n)$ on $n$ generators, let $\mathcal{T}_{\mathcal{V}}(n, m)$ be the set of homomorphisms $F(m) \to F(n)$, composed in the opposite order. This is a Lawvere theory. Here $\mathcal{T}_{\mathcal{V}}(n, 1) \cong F(n)$ is the set of $n$-ary terms modulo the identities of $\mathcal{V}$. The object $n$ is the product of $n$ copies of $1$ because $F(n)$ is the coproduct of $n$ copies of $F(1)$ in $\mathcal{V}$. Its category of models is equivalent to $\mathcal{V}$.

Non-example: the category $\aleph_0$ itself, not its opposite, has finite products, given by cartesian products, so the product of $n$ and $m$ is $n m$. It is not a Lawvere theory, since in a Lawvere theory every object is a power of one object $x$. Here $2$ and $3$ are not both powers of the same natural number.