# Epimorphism Preserves Distributivity

## 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 $+_2$.

$\blacksquare$