Epimorphism Preserves Identity

Theorem
Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structures.

Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an epimorphism.

Let $\left({S, \circ}\right)$ have an identity element $e_S$.

Then $\left({T, *}\right)$ has the identity element $\phi \left({e_S}\right)$.

Proof
Let $\left({S, \circ}\right)$ 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:

and:

The result follows because every element $y \in T$ is of the form $\phi \left({x}\right)$ 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