# Algebraic Functors and Left Adjoints

I'm currently taking a class on Categorical Logic, and we're just finishing our section on Lawvere Duality for algebraic theories.

As a quick remark, our profesor mentioned that every algebraic functor admits a left adjoint, and gave an example which feels wrong to me. Definitions and the example are below:

Recall a Lawvere Algebraic Theory $$\mathbb{A}$$ is a finite product category with objects $$A^n$$ for $$n \in \omega$$. $$A = A^1$$ is called the generator or universal model of $$\mathbb{A}$$.

The category of (set-valued) $$\mathbb{A}$$-models is $$\mathsf{Mod}(\mathbb{A}) = \mathsf{FP}(\mathbb{A}, \mathsf{Set})$$, the finite product preserving functors into Set.

Then a (finite product) functor $$F : \mathbb{A} \to \mathbb{B}$$ induces a functor $$F^* : \mathsf{Mod}(\mathbb{B}) \to \mathsf{Mod}(\mathbb{A})$$. That is, an $$\mathbb{A}$$-model in the (syntax) category $$\mathbb{B}$$ induces a (semantic) map from $$\mathbb{B}$$-models to $$\mathbb{A}$$-models. Such a functor $$F^*$$ is called Algebraic.

One can prove (though I have not seen the proof) that algebraic functors have left adjoints. As an example of this phenomenon, my professor said the following:

If $$\mathbb{G}$$ and $$\mathbb{R}$$ are the syntactic categories of groups and rings, respectively, then we have the map $$F : \mathbb{G} \to \mathbb{R}$$ which sends the generator to the generator. This map induces $$F^* : \mathsf{Ring} \to \mathsf{Group}$$, the functor taking a ring to its underlying abelian group.

He then remarked the guaranteed left adjoint $$G : \mathsf{Group} \to \mathsf{Ring}$$ is the group ring functor (I assume over $$\mathbb{Z}$$), however this doesn't sit well with me.

The "group ring" functor is left adjoint to the "group of units" functor, but adjoints are unique and so something must be wrong. Perhaps there is some alternative notion of group ring that makes this go through?

The questions, then, are:

• Am I correct in thinking that the group ring functor cannot be left adjoint to the underlying group functor?
• If so, then what is the guaranteed left adjoint? I cannot come up with a good candidate, and Google seems to be of little help in the matter.
• Are there references which will discuss these results? I am aware of Borceux's 3 volumes on the subject, but I am curious if there are other good sources.

You're correct, the left adjoint isn't the group ring construction. The underlying additive group functor $$\text{Ring} \to \text{Group}$$ splits as a composite

$$\text{Ring} \to \text{Ab} \to \text{Grp}$$

so its left adjoint splits as a composite going the other way. The left adjoint of $$\text{Ab} \to \text{Grp}$$ is abelianization, and the left adjoint of $$\text{Ring} \to \text{Ab}$$ is the tensor algebra functor. Their composite sends a group $$G$$ to the tensor algebra

$$T(A) = \bigoplus_{n \ge 0} A^{\otimes n}$$

of the abelianization $$A = G/[G, G]$$.

Similarly the left adjoint to the underlying additive group functor from commutative rings to abelian groups is given by taking the symmetric algebra

$$S(A) = \bigoplus_{n \ge 0} A^{\otimes n} / S_n.$$

• I don't know a reference for the construction but you can more or less do it "by hand." Freely adjoin the results of all new operations, then impose all new axioms. Oct 4, 2019 at 1:03

The free ring (with unit, noncommutative) on an abelian group $$A$$ is the tensor algebra $$\sqcup A^i$$, with multiplication given by concatenation. The left adjoint to the forgetful functor from rings to groups just composes this with abelianization. Left adjoints to algebraic functors are always free constructions: in this case, one simply adds in whatever products are needed. This is indeed not the group ring. One should specify that $$F$$ sends the multiplication in the syntax of groups to the addition in the syntax of rings, so that a connection to group rings, where the group structure is Incorporated via multiplication, wouldn't make sense.

Regarding the group ring, it cannot be the left adjoint of an algebraic functor, because it doesn't preserve retracts of finitely generated free objects. Indeed, $$\mathbb Z[X,X^{-1}]$$, the group ring of $$\mathbb{Z}$$, is not even a subring of any free ring, since free rings have finite groups of units. The group ring is, however, the left adjoint of an accessible functor, one living in the doctrine of locally (finitely) presentable categories, as it does preserve finitely presentable objects.

An excellent reference on algebraic categories is Algebraic Theories, by Adamek, Rosicky, and Vitale.

• The group ring functor is the left adjoint of the group of units functor. Is it not?
– user403337
Oct 4, 2019 at 1:52
• @Chris: yes, that's right. The group of units functor isn't algebraic, as defined in the OP (for the very simple reason that it doesn't preserve underlying sets). Oct 4, 2019 at 1:59
• Ahh. Thanks for the clarification @QiaochuYuan
– user403337
Oct 4, 2019 at 2:03
• @QiaochuYuan An algebraic functor preserves limits and sifted colimits, but it need not preserve the underlying set. Oct 4, 2019 at 4:55
• @Kevin: the OP defines an algebraic functor to be a functor between models of a Lawvere theory induced by a map of Lawvere theories. Presumably there are other notions of "algebraic" around. Oct 4, 2019 at 5:04