Epimorphism Preserves Distributivity

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Theorem

Let $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ be algebraic structures.

Let $\phi: R_1 \to R_2$ be an epimorphism.

If $\circ_1$ is left distributive over $+_1$, then $\circ_2$ is left distributive over $+_2$.
If $\circ_1$ is right distributive over $+_1$, then $\circ_2$ is right distributive over $+_2$.


Consequently, if $\circ_1$ is distributive over $+_1$, then $\circ_2$ is distributive over $+_2$.


That is, epimorphism preserves distributivity.


Proof

Throughout the following, we assume the morphism property holds for $\phi$ for both operations.


Left Distributivity

Suppose $\circ_1$ is left distributive over $+_1$. Then:

\(\displaystyle \map \phi x \circ_2 \paren {\map \phi y +_2 \map \phi z}\) \(=\) \(\displaystyle \map \phi x \circ_2 \map \phi {y +_1 z}\)
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {x \circ_1 \paren {y +_1 z} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {\paren {x \circ_1 y} +_1 \paren {x \circ_1 z} }\) as $\circ_1$ is left distributive over $+_1$
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {x \circ_1 y} +_2 \phi \paren {x \circ_1 z}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\map \phi x \circ_2 \map \phi y} +_2 \paren {\map \phi x \circ_2 \map \phi z}\)

So $\circ_2$ is left distributive over $+_2$.

$\blacksquare$


Right Distributivity

Suppose $\circ_1$ is right distributive over $+_1$. Then:

\(\displaystyle \paren {\map \phi x +_2 \map \phi y} \circ_2 \map \phi z\) \(=\) \(\displaystyle \map \phi {x +_1 y} \circ_2 \map \phi z\)
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {\paren {x +_1 y} \circ_1 z}\)
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {\paren {x \circ_1 z} +_1 \paren {y \circ_1 z} }\) as $\circ_1$ is right distributive over $+_1$
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {x \circ_1 z} +_2 \map \phi {y \circ_1 z}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\map \phi x \circ_2 \map \phi z} +_2 \paren {\map \phi y \circ_2 \map \phi z}\)

So $\circ_2$ is right distributive over $+_2$.

$\blacksquare$


Distributivity

If $\circ_1$ is distributive over $+_1$, then it is both right and left distributive over $+_1$.

Hence from the above, $\circ_2$ is both right and left distributive over $+_2$.

That is, $\circ_2$ is distributive over $+_2$.

$\blacksquare$