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 $R_1$, $\paren {x +_1 y}, \paren {y +_1 z}, \paren {x \circ_1 y }, \paren {y \circ_1 z}, \paren {x \circ_1 z}, \paren { x \circ_1 \paren {y +_1 z} }, \paren { \paren {x +_1 y} \circ_1 z } \in \Dom \phi$.

This theorem requires a proof. In particular: The domain of the epimorphism must be verified for a valid proof, since the algebraic structure are not assumed to be closed under its binary operation. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Left Distributivity

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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 $+_1$, $+_2$, $\circ_1$ 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:

\(\ds \map \phi x \circ_2 \paren {\map \phi y +_2 \map \phi z}\)

\(=\)

\(\ds \map \phi x \circ_2 \map \phi {y +_1 z}\)

as $\circ_2$ is left distributive over $+_2$

\(\ds \)

\(=\)

\(\ds \map \phi {x \circ_1 \paren {y +_1 z} }\)

\(\ds \)

\(=\)

\(\ds \map \phi {\paren {x \circ_1 y} +_1 \paren {x \circ_1 z} }\)

as $\circ_1$ is left distributive over $+_1$

\(\ds \)

\(=\)

\(\ds \map \phi {x \circ_1 y} +_2 \phi \paren {x \circ_1 z}\)

\(\ds \)

\(=\)

\(\ds \paren {\map \phi x \circ_2 \map \phi y} +_2 \paren {\map \phi x \circ_2 \map \phi z}\)

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

$\blacksquare$

Right Distributivity

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

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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:

\(\ds \paren {\map \phi x +_2 \map \phi y} \circ_2 \map \phi z\)

\(=\)

\(\ds \map \phi {x +_1 y} \circ_2 \map \phi z\)

as $\circ_2$ is right distributive over $+_2$

\(\ds \)

\(=\)

\(\ds \map \phi {\paren {x +_1 y} \circ_1 z}\)

\(\ds \)

\(=\)

\(\ds \map \phi {\paren {x \circ_1 z} +_1 \paren {y \circ_1 z} }\)

as $\circ_1$ is right distributive over $+_1$

\(\ds \)

\(=\)

\(\ds \map \phi {x \circ_1 z} +_2 \map \phi {y \circ_1 z}\)

\(\ds \)

\(=\)

\(\ds \paren {\map \phi x \circ_2 \map \phi z} +_2 \paren {\map \phi y \circ_2 \map \phi z}\)

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 $+_2$.

$\blacksquare$

Warning

Note that this result is applied to epimorphisms.

For a general homomorphism which is not surjective, nothing definite can be said about the behaviour of the elements of its codomain which are not part of its image.