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