Construction of free objects in an arbitrary variety using only the HSP operations

Question

According to Birkhoff's HSP theorem, a class of algebraic structures is a variety if and only if it is closed under the taking of homomorphic images, subalgebras, and arbitrary products. It is also known that any category corresponding to a variety has free objects. This suggests to me that there should be some general construction for free objects, using only the operations of taking homomorphic images, subalgebras, and products, which works for any variety. Does such a construction exist?

Answer 1

The proof of the Freyd adjoint theorem gives exactly such a construction. Namely, fix some set $X'$ which is either $X$ if $X$ is infinite, or $\mathbb{N}$ if $X$ is finite. Now, take the product of $A^{A^X}$ for all algebras $A$ on a subset of $X'$. For each $x \in X$, we have the element of $\prod_A A^{A^X}$ whose $A$-element is the projection $\varphi \in A^X \mapsto \varphi(x) \in A$. The subalgebra of $\prod_A A^{A^X}$ generated by these elements will be a representation of the free algebra on $X$. (The homomorphic image condition will presumably come into play in proving that this is actually a left adjoint.)

Answer 2

You can see the essence of this construction in George Bergman's Invitation to General Algebra and Universal Constructions, in Section 3.3, "Free groups as subgroups of big enough direct products." As Daniel Schepler mentions, the construction is based on the proof of Freyd's Adjoint Functor Theorem.

Assume you have a fixed signature $\Omega$, let $\mathfrak{V}$ be a variety of $\Omega$-algebras, and let $X$ be a set.

Let $S$ be a set of cardinality $\max(|X|,\aleph_0)$, and let $I$ be a set of ordered pairs $(A,f)$, where $A$ is an algebra of $\mathfrak{V}$ with underlying set contained in $S$, and $f\colon X\to A$ is a function from $X$ to the underlying set of $A$, such that for every algebra $B\in\mathfrak{V}$ and every function $g\colon X\to B$, there exists a unique pair $(A,f)$ and a unique isomorphism $\psi\colon B\to A$ such that $f=\psi\circ g$. (Restricting to underlying sets contained in $S$ is done to ensure this family is a set). Now let $P$ be the product $$P = \prod_{(A,f)\in I} A.$$ Given $x\in X$ and $(A,f)\in I$, we have an element $f(x)\in A$. Thus, each $x\in X$ gives a unique element of $P$, $(f(x))_{(A,f)\in I}$. Let $F$ be the subalgebra of $P$ generated by these elements: $$F = \bigl\langle (f(x))_{(A,f)\in I}\bigr\rangle.$$ Let $i\colon X\to F$ be the map that sends $x\in X$ to $(f(x))_{(A,f)\in I}$.

I claim that $(F,i)$ is a free object on $X$ in $\mathfrak{V}$. It lies in $\mathfrak{V}$, since it is a subobject of a product of objects in $\mathfrak{V}$. Let $B\in\mathfrak{V}$, and let $j\colon X\to B$ be any map. By choise of $I$, there exists a unique index $(A,f)\in I$ and a unique isomorphism $\psi\colon B\to A$ such that $f=\psi\circ j$. Now consider the map $\pi_{(A,f)}\colon P\to A$ given by the projection onto the $(A,f)$th coordinate. Then we have that for each $x\in X$, $$\psi^{-1}\circ\pi_{(A,f)}(i(x)) = \psi^{-1}\circ f(x) = \psi^{-1}\circ\psi\circ j(x)=j(x).$$ Thus, $\Phi=\psi^{-1}\circ\pi_{(A,f)}\colon F\to B$ has the property that $\Phi\circ i = j$.

Moreover, if $\Theta\colon F\to B$ is such that $\Theta\circ i = j$, then for each $x\in X$ we have that $$\Theta(i(x)) = j(x) = \Phi(i(x)),$$ so $\Theta$ and $\Phi$ agree on all of $i(X)$. Since $F=\langle i(X)\rangle$, then $\Theta=\Phi$, as two morphisms that agree on a generating set must be equal to each other. Thus, $\Phi$ is the unique morphism $F\to B$ such that $\Phi\circ i = j$.

This proves that $(F,i)$ has the universal property of the free object on $X$, hence is the free object on $X$.