Epimorphism Preserves Distributivity

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Theorem

Let $\left({R_1, +_1, \circ_1}\right)$ and $\left({R_2, +_2, \circ_2}\right)$ be algebraic structures.

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


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:

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

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

$\blacksquare$


Right Distributivity

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

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

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

$\blacksquare$


Distributive

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 $+_2$.

$\blacksquare$