# Concrete functors with left adjoint

Let $$R$$ be a commutative ring. The category $$R$$-Mod of modules over $$R$$ has a natural faithful functor into Set sending a $$R$$-module to its underlying set and a $$R$$-module morphism to its underlying function. The same is true for Grp and for many other categories. Furthermore all of these forgetful functors have a left adjoint, which means there are free objects in these categories with respect to those forgetful functors.

Are there exotic concrete category structures with a left adjoint on any of these categories? How are the free objects wrt those structures like? By concrete category i mean a faithful functor from a category to Set, which we call its forgetful functor.

• What's an "exotic concrete category structure" on a category? Commented Feb 10 at 21:40
• @BenSteffan, I would imagine that "exotic" concrete categories would not be "naturally" categories of things with underlying sets, but would have some crazy functor to Sets...? Perhaps the questioner can clarify. (I myself know of no such example.) Commented Feb 10 at 21:43
• @paulgarrett ...that's why I'm asking! Commented Feb 10 at 21:44
• @BenSteffan :) ... Conceivably, some "champion of set theory as a basis for all of mathematics" could contrive set-theoretic constructions for nearly anything. Maybe not so natural, though. As in "is the number 1 really $\{\{\},\{\{\}\}$? :) Commented Feb 10 at 21:48
• Yes, what paul garrett said. Instead of having the usual forgetful functor, having some other. Commented Feb 11 at 0:49

Let $$C$$ be a category with set-indexed copowers (meaning that if $$A$$ is any object in $$C$$ and $$I$$ is any set, then family $$(A_i)_{i\in I}$$ with $$A_i = A$$ for all $$i\in I$$ has a coproduct, denoted $$A^{(I)}$$).

A functor $$\newcommand{\Set}{\mathsf{Set}}\newcommand{\Hom}{\mathrm{Hom}}U\colon C\to \Set$$ has a left adjoint if and only if $$U$$ is representable. Indeed, if $$F\dashv U$$, then $$U(-)\cong \Hom_\Set(*,U(-))\cong \Hom_C(F(*),-)$$, so the object $$F(*)$$ represents $$U$$. Conversely, if $$U$$ is representable by an object $$A$$, then we have $$\Hom(A^{(I)},-)\cong (\Hom(A,-))^I \cong U(-)^I \cong \Hom_\Set(I,U(-))$$, so the functor $$F(I) = A^{(I)}$$ is left-adjoint to $$U$$.

Now a representable functor $$\Hom_C(A,-)\colon C\to \Set$$ is faithful if and only if $$A$$ is a separator (also known as a generator) for $$C$$. It could be useful to rephase the definition of separator as follows. For an object $$A$$, the following are equivalent:

1. $$A$$ is a separator, i.e., for any pair of distinct arrows $$g,h\colon B\to C$$, there exists an arrow $$f\colon A\to B$$ such that $$gf \neq hf$$.
2. For every object $$B$$ there exists a set $$I$$ and an epimorphism $$A^{(I)}\to B$$.
3. For every object $$B$$, the canonical map $$A^{(\Hom(A,B))}\to B$$ coming from the universal property of the coproduct is an epimorphism.

So to answer your question: The standard concrete structure on an algebraic category like the ones you mention comes from the free object on one generator ($$R$$ in the case of $$R\mathsf{-Mod}$$, $$\mathbb{Z}$$ in the case of $$\mathsf{Grp}$$) which is a separator and represents the underlying set functor. The "exotic" concrete structures which admit left adjoints arise in the same way from the other separators in the category.

For an explicit example: in $$\mathsf{Ab}$$, the object $$\mathbb{Z}^n$$ is a separator for any $$n\geq 1$$. The functor it represents maps $$A$$ to the underlying set of $$A^n$$.

A bit less trivially: in $$\mathsf{Grp}$$, the coproduct $$\mathbb{Z}\coprod C_2$$ is a separator. The functor it represents maps $$G$$ to $$\{(g,h)\in G^2 \mid h\text{ has order \leq 2}\}$$.