Let $X$ denote an object of a finite-product category $\mathbf{C}$. Then $X$ generates a finite-product subcategory $\mathrm{Lawv}(X),$ and $\mathrm{Lawv}(X)$ is automatically a Lawvere theory.

Examples.

• Viewing $2$ as an object of $\mathbf{Set}$, we see that the arrows of $\mathbf{Lawv}(2)$ are "truth tables." Hence $\mathbf{Lawv}(2)$ is the Lawvere theory of Boolean algebras.

• Let $\mathbb{K}$ denote a field. Write $U_{\mathrm{Mod}}(\mathbb{K})$ for the underlying $\mathbb{K}$-module. Then $\mathbf{Lawv}(U_{\mathrm{Mod}}(\mathbb{K}))$ is the Lawvere theory of $\mathbb{K}$-modules.

Question. Is it true that for all Lawvere theories $\mathsf{T}$, there exists a Lawvere theory $\mathsf{S}$ and an $\mathsf{S}$-algebra $X$ such that $\mathrm{Lawv}(X) \cong \mathsf{T}$?