# Composition of Ring Endomorphisms is Ring Endomorphism

## Theorem

Let $R$ be a set.

Let:

$\left({R, +_1, \circ_1}\right)$
$\left({R, +_2, \circ_2}\right)$
$\left({R, +_3, \circ_3}\right)$

be rings.

Let:

$\phi: \left({R, +_1, \circ_1}\right) \to \left({R, +_2, \circ_2}\right)$
$\psi: \left({R, +_2, \circ_2}\right) \to \left({R, +_3, \circ_3}\right)$

Then the composite of $\phi$ and $\psi$ is also a (ring) endomorphism.

## Proof

A ring endomorphism is a ring homomorphism $f$ from a set to itself.

That is:

$\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\phi}\right)$
$\operatorname{Dom} \left({\psi}\right) = \operatorname{Cdm} \left({\psi}\right)$

From Composition of Ring Homomorphisms is Ring Homomorphism, $\psi \circ \phi$ is a ring homomorphism.

By definition of composition of mappings:

$\operatorname{Cdm} \left({\phi}\right) = \operatorname{Dom} \left({\psi}\right)$

Thus:

$\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\psi}\right) = R$

and so $\psi \circ \phi$ is a ring endomorphism.

$\blacksquare$