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?

  • $\begingroup$ 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. $\alpha: L\to L$. $\endgroup$ Aug 20, 2013 at 10:47
  • 6
    $\begingroup$ An algebraic theory in which every term is a homomorphism is called commutative: ncatlab.org/nlab/show/commutative+algebraic+theory $\endgroup$ Aug 20, 2013 at 12:33
  • $\begingroup$ @Tom: This is a full answer. $\endgroup$ Aug 20, 2013 at 13:39
  • $\begingroup$ Hint taken, Martin! $\endgroup$ Aug 20, 2013 at 16:04

2 Answers 2


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 $\alpha_S(x)=\widehat{x}\circ a$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.