In Mac Lane's Catergories for the Working Mathematcian text, he introduces the concepts of universal algebra to explain why the forgetful functors $\textbf{Grp} \to \textbf{Set}$, $\textbf{Ab} \to \textbf{Set}$, etc., all have left adjoints (page 124).

He states that given an algebraic system ($\Omega$, $E$) where

  • $\Omega$ is the set of operators

  • $E$ is the set of identities (which are pairs of $\Omega$) (This is wrong. See answer.)

one can form the category of $(\Omega, E)$-$\textbf{Alg}$ containing all algebras of type $(\Omega, E)$ with morphisms preserving the operations.

Basically, he says (but doesn't show) that one can construct a left adjoint of the forgetful functor $U: (\Omega, E)$-$\textbf{Alg} \to \textbf{Set}$ for any algebraic system $(\Omega, E)$.

My Questions:

  1. $\textbf{Comp Bool}$, the category of complete Boolean algebras, is a category of an algebraic system, whose forgetful functor to $\textbf{Set}$ has no left adjoint (because Solovay's theorem shows that an infinite set cannot produce a free complete Boolean Algebra). How does this not contradict his claim?

  2. How does one construct, for an algebraic system $(\Omega, E)$, the left adjoint of the forgetful functor $U: (\Omega, E)$-$\textbf{Alg} \to \textbf{Set}$ (the general "free functor")?

A reference would suffice too, I'm just having a hard time finding this result online. Also, feel free to tell me if my understanding of algebraic systems is wrong; I've only seen them from Mac Lane at this point.

This is a similar question, but I'm looking for a proof that the functor exists to begin with. The notion of free algebra functor

  • $\begingroup$ You claim that the category of complete Boolean algebras is the category of an algebraic system. Can you actually construct $\Omega,E$ so that $(\Omega,E)\text{-Alg}$ coincides with this category? $\endgroup$
    – Z. A. K.
    Jan 17, 2020 at 1:42
  • $\begingroup$ @Z.A.K. I think I see what you're saying. Is the added hypothesis of completeness on the boolean algebras incapable of being modeled by $\Omega$ and $E$ alone? $\endgroup$
    – trujello
    Jan 17, 2020 at 1:49
  • $\begingroup$ That's right! There is no set of identities that hold precisely among the operators complete Boolean algebras - and this is true in more generality, see e.g. the following answer: math.stackexchange.com/questions/3419751/… $\endgroup$
    – Z. A. K.
    Jan 17, 2020 at 2:00

1 Answer 1


Question 1

As we have discussed in the comments, $\textbf{Comp Bool}$ does not form a the category of any algebraic system. More generally, the theory of complete Boolean algebras is not even first-order, much less algebraic.

Question 2

Consider a set $X$. To construct the value of $F(X)$ for an equational theory $(\Omega, E)$, we first construct a term algebra $T(X)$ of $\Omega$: this is the set formed by all valid expressions obtained by treating elements of $X$ as variables, and elements of $\Omega$ as $n$-ary operations (you'll see that Mac Lane assigns to each element $\omega \in \Omega$ a natural number $n$, called the arity of $\omega$; for an arbitrary set of variables).

Now, noticing that the elements of $E$ are not pairs from $\Omega$ as you claim, but pairs of "derived operators" (the construction of derived operators is closely related to the term algebra construction over a countable set of variables), we can consider the least equivalence relation $\sim_E$ on $T(X)$ that satisfies the identities in $E$. The value of $F(X)$ can then be defined as the quotient $T(X)/\sim_E$. One needs to check that this construction indeed yields a functor, and verify the adjointness conditions. It's a useful exercise to do this for groups first, following along with any standard Abstract Algebra textbook, to see that this gets us the usual construction of free groups as reduced words.

For a brief reference on the construction, you may wish to consult Section 2 of On the Construction of Free Algebras for Equational Systems by Fiore and Hur.

  • $\begingroup$ Wow, this is super helpful. Thanks so much as this is what I need! $\endgroup$
    – trujello
    Jan 17, 2020 at 2:40

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