# Why fully invariant congruence is an algebraic closure operator?

If we have an algebra $A$ of type $F$ then congruence of fully invariant is an algebraic closure structure operator on $A\cdot A$.

Actually it's in Universal Algebra Sankappanavar page $100$ (Lemma $14.4$).

And specially I'm asking why the fully invariant congruence is an algebraic closure operator?

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Sorry, can you spell out what do you mean by "congruence of fully invariant"? –  Berci Dec 15 '12 at 19:56
@Berci: The definition that MohammadSadegh is using is: A congruence $\theta$ on an algebra $\mathbf{A}$ is fully invariant if for every endomorphism $\alpha$ on $\mathbf{A}$, $$\langle a,b\rangle\in\theta\implies\langle\alpha a,\alpha b\rangle\in\theta\;.$$ –  Brian M. Scott Dec 15 '12 at 20:10
Yes, yes, thanks, just the composition of the sentence was not really clear.. –  Berci Dec 15 '12 at 20:15
That's exactly the endomorphism which Mr Brian mentioned. –  MohammadSadegh YazdanParast Dec 16 '12 at 6:23
@Brian: Thank you Mr. –  MohammadSadegh YazdanParast Dec 16 '12 at 6:23

Let $\mathbf{B}$ be the algebra $\mathbf{A}\times\mathbf{A}$ together with the new fundamental operations listed in the proof in Burris & Sankappanavar:

\begin{align*} \langle a,a\rangle&\qquad\text{for }a\in A\\ s\big(\langle a,b\rangle\big)=\langle b,a\rangle&\\ t\big(\langle a,b\rangle,\langle c,d\rangle\big)=\begin{cases}\langle a,d\rangle\\ \langle a,b\rangle \end{cases}&\qquad\begin{array}{l}\text{if }b=c\\\text{otherwise}\end{array}\\ e_\sigma\big(\langle a,b\rangle\big)=\langle\sigma a,\sigma b\rangle&\qquad\text{for }\sigma\text{ and endomorphism of }\mathbf{A}\;. \end{align*}

The key step is the assertion that $\theta$ is a fully invariant congruence on $\mathbf{A}$ iff $\theta$ is a subuniverse of of $\mathbf{B}$.

Let $\theta$ be a congruence on $\mathbf{A}$, and note that $\theta\subseteq A\times A$. If $\theta$ is a subuniverse of $\mathbb{B}$, then by definition $\theta$ is closed under the fundamental operations of $\mathbf{B}$. In particular, $\theta$ is closed under $e_\sigma$ for each endomorphism $\sigma$ of $\mathbf{A}$, which is exactly what it means for $\theta$ to be fully invariant.

Now suppose that $\theta$ is fully invariant. By the definition of congruence (Definition 5.1) $\theta$ is an equivalence relation, so it’s closed under the first three operations listed above. Definition 5.1 also requires that $\theta$ be closed under the fundamental operations of $\mathbf{A}$. Finally, $\theta$ is closed under the new operations $e_\sigma$ because by hypothesis it’s fully invariant. $\dashv$

It follows from Theorem 3.2 that $\Theta_{\text{FI}}$ is an algebraic closure operator: for any $S\subseteq A\times A$,

\begin{align*} \Theta_{\text{FI}}(S)&=\bigcap\{\theta\in\operatorname{Con}_{\text{FI}}(\mathbf{A}):S\subseteq\theta\}\\ &=\bigcap\{B\subseteq A\times A:S\subseteq B\text{ and }B\text{ is a subuniverse of }\mathbf{B}\}\\ &=\operatorname{Sg}(S)\;. \end{align*}

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@MohammadSadeghYazdanParast: You’re very welcome. –  Brian M. Scott Dec 16 '12 at 18:13
1. Consider the algebraic structure on $A\times A$ with the following operations:

• the original $\mu$ operations acting simultaneously on both coordinates (i.e. so far this is the usual structure on $A\times A$): $\ \underset i\mu (a_i,b_i):= (\underset i\mu a_i, \underset i\mu b_i)$
• the unary operation guaranteeing symmetry of the subalgebras: $(a,b)\mapsto (b,a)$
• similarly, for transitiveness, the binary operation $*$ which acts as $(a,b)*(b,c)=(a,c)$ and, say, $(a,b)*(c,d)=(a,b)$ if $b\ne c$
• for each $a\in A$, a constant operation $(a,a)$ for reflexivity of subalgebras (So far, this is an algebraic structure on the set $A\times A$ such that its subalgebras are exactly the congruences.)
• Now, for each endomorphism $\alpha:A\to A$, introduce one more unary operation, sending $(a,b)\mapsto (\alpha(a),\alpha(b))$.
2. Anyway, there can be general conditions said to the collection $\mathcal C$ of the required 'closed sets', in order that it determines an algebraic closure operator by defining the closure $cl\, S$ of a set $S$ as $\bigcap\{C\in\mathcal C\mid S\subseteq C\}$. And these conditions are:

• $\mathcal C$ is closed under intersections (of arbitrary many elements)
• some finiteness condition, something corresponding to $cl\, S=\bigcup\{cl\,H \mid H\subseteq S\text{ finite}\}$.
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That was useful my friend. –  MohammadSadegh YazdanParast Dec 16 '12 at 6:36