# Terms of an algebraic theory that act as homomorphisms

Say you have some kind of "algebraic" category $A$ with a forgetful functor $U : A \to \mathbf{Set}$ which has a left adjoint $F : \mathbf{Set} \to A$. The natural transformations $U \to U$ can be interpreted as the terms of the algebraic theory. For example, if $A=\mathbf{CRing}$, these are just polynomials. Suppose we consider the map $\Phi : \mathrm{Nat}(id_A,id_A) \to \mathrm{Nat}(U,U)$ that simply takes a natural transformation $\eta : id_A \to id_A$ and performs horizontal composition with the identity natural transformation $U \to U$. What is the interpretation of the image of $\Phi$? That is, what are the terms that actually act as morphisms of $A$?

Why do $\mathbf{Ab}$ and related categories enjoy the property that $\Phi$ is surjective, i.e. every term acts as a morphism?

• Nat. tr. $U\to U$ are in (contravariant) bijection with nat. tr. $\alpha: L\to L$ and for any set $S$ a morphisms $A: L(S) \to L(S)$ is identified to family of terms $(A_s\in L(S))_{s\in S}$. The problem is how a collection of families $(A_s\in L(S))_{s\in S}$ correspond to a nat.tr. $\alpha: L\to L$. Aug 20 '13 at 10:47
• An algebraic theory in which every term is a homomorphism is called commutative: ncatlab.org/nlab/show/commutative+algebraic+theory Aug 20 '13 at 12:33
• @Tom: This is a full answer. Aug 20 '13 at 13:39
• Hint taken, Martin! Aug 20 '13 at 16:04

An algebraic theory in which every term is a homomorphism is called commutative: ncatlab.org/nlab/show/commutative+algebraic+theory. I'm not sure that's quite what you were asking, because here "term" means "term in any number of variables", that is, natural transformation $U^n \to U$ for any $n$ (not just $1$). For instance, in the theory of abelian groups, $+: A \times A \to A$ is a homomorphism for any abelian group $A$. But commutative theories are certainly a useful class to consider, and a lot has been written about them.

Given two adjunctions $<F, G>, <F', G'>: \mathscr{B}\to \mathscr{A}$ from standard properties follow the natural isomorphism $Nat(F, F') \cong Nat(1, G\circ F') \cong Nat(G', G)$.

In our case follow that there is a isomorphism between the monoid of the natural transformations $U \Rightarrow U$ and the dual of the monoid of natural transformations $L \Rightarrow L$.

Let $\alpha : L \Rightarrow L$, given a set $S$ we have the commutative diagram

$\begin{array}{ccc} L(1) & \xrightarrow{\alpha_1 } & L(1) \\ \widehat{x} \downarrow & &\downarrow \widehat{x} \\ L(S)& \xrightarrow{\alpha_S } & L(S) \end{array}$

where $x\in X$ is identified by the unique morphism $\widehat{x}: L(1) \to L(S)$ such that $\widehat{x}(1)=x$.

I claim that $\alpha \mapsto \alpha_1$ is a bijection between the natural transformations $\alpha: L \Rightarrow L$ and the morphisms $a: L(1) \to L(1)$ (i.e. the set $L(1)$),

The set of morphisms $\widehat{x}: L(1) \to L(S)$ is a epimorphic family, then from the diagram this map is injective, given $a: L(1) \to L(1)$, from the diagram define $\alpha_S: L(S) \to L(S)$ as the (unique) morphisms such that $\alpha_S(x)=\widehat{x}\circ a$.