Epimorphism Preserves Distributivity

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.

It remains to be shown that for non-empty $R_1$, $\paren {x +_1 y}, \paren {y +_1 z}, \paren {x \circ_1 y }, \paren {y \circ_1 z}, \paren {x \circ_1 z}, \paren { x \circ_1 \paren {y +_1 z} }, \paren { \paren {x +_1 y} \circ_1 z } \in \Dom \phi$.

Left Distributivity
Suppose $R_1$ is the empty set.

It follows from the definition of an epimorphism that $\phi$ is a surjective homomorphism

By Empty Mapping to Empty Set is Bijective, the empty map is bijective By definition of bijection, the empty map is an epimorphism.

Therefore, suppose $R_1$ is the empty map, which is indeed an epimorphism.

By Image of Empty Set is Empty Set, $R_2$ is also the empty set.

It follows from the definition of the homomorphism that the binary operations $\circ_1$ and $\circ_2$ are also the empty map.

If $\circ_1$ is left distributive over $+_1$, then it is vacuously true that $\circ_2$ is left distributive over $+_2$, as required.

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be an epimorphism.

Suppose $R_1$ is non-empty.

Then:

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

Right Distributivity
Suppose $\circ_1$ is right distributive over $+_1$.

Suppose $R_1$ is the empty set.

By the similar reasoning shown for left distributivity, it follows that if $\circ_1$ is left distributive over $+_1$, then it is vacuously true that $\circ_2$ is left distributive over $+_2$, as required.

Suppose $R_1$ is non-empty.

Then:

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

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 $+_2$.