If $(\alpha,\beta)$ is the factor pair congruences of algebra $\mathbb{A},$ ia $(\forall \gamma\in ConA)\alpha\circ\gamma=\gamma\circ\beta?$

Question

Let $\mathbb{A}$ be an algebra such that $ConA$ is the distributive lattice. If $(\alpha,\beta)$ is the factor pair congruences of algebra $\mathbb{A},$ prove that $(\forall \gamma\in ConA)\alpha\circ\gamma=\gamma\circ\beta.$

Answer 1

It isn't true.
Just take $\gamma$ to be $\{(a,a):a\in A\}$, that gives $$\alpha = \alpha \circ \gamma = \gamma \circ \beta = \beta,$$ which clearly doesn't hold.

Answer 2

Edited Question. Let $\mathbb{A}$ be an algebra such that ${\rm Con}(\mathbb{A})$ is a distributive lattice. If $(\alpha,\beta)$ is a factor pair of congruences of $\mathbb{A}$, prove that $\alpha$ permutes with every $\gamma$ in ${\rm Con}(\mathbb{A})$. (That is, $\alpha\circ\gamma=\gamma\circ\alpha$ for every $\gamma\in {\rm Con}(\mathbb{A})$.)

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Choose any $\gamma$ in ${\rm Con}(\mathbb{A})$. You can prove that $\alpha$ permutes with $\gamma$ in three steps.

(1) Show that $\alpha$ permutes with $\alpha\cap\gamma$.

(2) Show that $\alpha$ permutes with $\beta\cap\gamma$.

(3) Show that $\alpha$ permutes with $(\alpha\cap\gamma)\vee(\beta\cap \gamma) = \gamma$.

For (1), use the fact that comparable equivalence relations permute.

For (2), use the facts that $\alpha$ permutes with $\beta$ and that ${\rm Con}(\mathbb{A})$ is distributive.
(Some extra details: Choose $(x,z)\in \alpha\circ (\beta\cap\gamma)$. There is some $y$ such that $(x,y)\in\alpha$ and $(y,z)\in\beta\cap\gamma$. Since $(x,z)\in\alpha\circ\beta=\beta\circ\alpha$ there exists $w$ such that $(x,w)\in\beta$ and $(w,z)\in\alpha$. Now $$(x,w)\in \beta\cap(\alpha\vee(\beta\cap\gamma))= (\beta\cap\alpha)\vee (\beta\cap\gamma) = \beta\cap\gamma. $$ Since $x\;(\beta\cap\gamma)\;w\;\alpha\;z$ we have $(x,z)\in (\beta\cap\gamma)\circ\alpha$, as desired.)

For (3), argue that if $\alpha$ permutes with two equivalence relations, then it permutes with their join.