Epimorphism Preserves Identity

Theorem
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 $\map \phi {e_S}$.

Proof
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.

Then:
 * $\forall x \in S: x \circ e_S = x = e_S \circ x$

Thus,
 * $\forall x \in S: x \circ e_S, e_S \circ x \in \Dom \phi$.

Hence:

and:

The result follows because every element $y \in T$ is of the form $\map \phi x$ with $x \in S$.

Also see

 * Group Homomorphism Preserves Identity


 * Epimorphism Preserves Associativity
 * Epimorphism Preserves Commutativity
 * Epimorphism Preserves Inverses


 * Epimorphism Preserves Semigroups
 * Epimorphism Preserves Groups


 * Epimorphism Preserves Distributivity