Epimorphism Preserves Properties
Theorem
Epimorphism Preserves Associativity
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\circ$ be an associative operation.
Then $*$ is also an associative operation.
Epimorphism Preserves Commutativity
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\circ$ be a commutative operation.
Then $*$ is also a commutative operation.
Epimorphism Preserves Distributivity
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.
Epimorphism Preserves Identity
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ have an identity element $e_S$.
Then $\struct {T, *}$ has the identity element $\map \phi {e_S}$.
Epimorphism Preserves Inverses
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Let $x^{-1}$ be an inverse element of $x$ for $\circ$.
Then $\map \phi {x^{-1} }$ is an inverse element of $\map \phi x$ for $*$.
That is:
- $\map \phi {x^{-1} } = \paren {\map \phi x}^{-1}$
Epimorphism Preserves Semigroups
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ be a semigroup.
Then $\struct {T, *}$ is also a semigroup.
Epimorphism Preserves Groups
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ be a group.
Then $\struct {T, *}$ is also a group.
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.