3.9 KiB
title: Counterexamples of algebraic theories tags: category theory, algebra date: 2020-01-21 00:00
An algebraic theory T is given by a collection of operations with
given arities, and a number of equations of the form $\Gamma \mid t =
s$ between terms of T. Here t and s can contain free variables
from \Gamma. A model M for an algebraic theory T is an
underlying set (call it M as well) which interprets all the
operations, and for which all the equations of T hold. A morphism of
T-models is a function of the underlying sets that presevres the
interpretation of all the operations.
The category Mod_T of T-models comes with a forgetful functor $U :
Mod_T \to Set$. It has a left adjoint F : Set \to Mod_T which
constructs a free T-model on the underlying set.
The category Mod_T and the functor U always satisfies a number of
properties that can be used for showing that certain theories are not
algebraic, i.e. that certain categories do not arise as models of
algebraic theories.
Conservativity of the forgetful functor
One of the properties of U is that it reflects isomorphisms: if $f :
M \to M' is a morphism of $T-models, and f has a set-theoretic
inverse g (i.e. g : M' \to M such that U(f) \circ g = 1 and $g
\circ U(f) = 1$), then f is an isomorphism of models.
This can be easily observed for monoids: let f be a group
homomorphism and let g be it's set theoretic inverse. Then $g(ab) =
g(f(g(a)) f(g(b))) = g (f (g(a)g(b))) = g(a)g(b)$.
Posets. A discrete two-element poset X = \\{0,1\\} can be embedded
into the poset Y = \\{0, 1\\} with 0 \leq_Y 1. Moreover, on a
set-theoretic level, this injection is just an identity function and
hence is an isomorphism. If the theory of posets were algebraic, then
X and Y were isomorphic as posets, which is not the case.
Lattice of submodels
Up to isomorphism, a subobject of a T-model M is a T-model M'
such that M' \subseteq M and the interpretation of all the
operations on M' coincides with the interpretations in M.
Like in many categories, subobjects for a given model form a lattice.
The meet of submodels is just a set-theoretic intersection (can be
computed from the pullback of one submodel along the other). However,
the join of two submodels may not coincide with the set-theoretic
union: if we have two submodels M_1, M_2, and a binary operation f
in the theory, then the set-theoretic union M_1 \cup M_2 will not
contain \bar{f}(x, y) for x \in M_1, y \in M_2 and \bar{f} the
interpretation of f in M.
Given a set X \subseteq M we define the closure of X to be the
smallest submodel \bar{X} \subseteq M that contains X. Explicitly,
\bar{X} = \\{ \bar{f}(x_1, \dots, x_n) \in M \mid x_1, \dots, x_n \in X, f \in T \\}
where \bar{f} is a M-interpretation of an n-ary operation f from T.
\bar{X} behaves like the closure operation and satisfies all the expected properties.
Using the closure operation we can define the join of two submodels M_1, M_2 \subseteq M as
M_1 \vee M_2 = \overline{M_1 \cup M_2}This can be generalized to unions over arbitrary families \bigvee_{i \in \mathcal{I}} M_i.
Finally, note the following: if M_i, i \in \mathcal{I} is a directed family, then \bigvee_{i \in \mathcal{I}} M_i = \bigcup_{i \in \mathcal{I}} M_i.
If at some point we add \bar{f}(a, b) to the join, with a \in M_a, b \in M_b, then there is some M_c that includes both M_a and M_b; this submodel M_c already contains \bar{f}(a, b).
Now we are ready to give another counterexample.
Complete lattices. The theory of complete lattices in not algebraic.
Consider the completion \omega+1 of \omega with the top element \infty.
Consider a directed family D = \\{ \\{ 0, \dots, n \\} \mid n \in \mathbb{N} \\}.
The set-theoretic union \bigcup D is just the natural numbers (not a complete lattice), but the join \bigvee D is \omega+1.