Epimorphism Preserves Identity

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

If $\circ$ has an identity $e_S$, then $\phi \left({e_S}\right)$ is the identity for $*$.

Proof
Let $\left({S, \circ}\right)$ be an algebraic structure in which $\circ$ has an identity $e_S$.

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

The result follows directly from the morphism property of $\circ$ under $\phi$:

Warning
Note that this result is applied to epimorphisms. For a general homomorphism which is not surjective, we can say nothing definite about the behaviour of the elements of its codomain which are not part of its image.

However, also see: for when the algebraic structure is actually a group.
 * Group Homomorphism Preserves Identity

Also see

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


 * Epimorphism Preserves Semigroups
 * Epimorphism Preserves Groups


 * Epimorphism Preserves Distributivity