# Composite of Homomorphisms is Homomorphism/Algebraic Structure

## Theorem

Let:

$\struct {S_1, \otimes_1, \otimes_2, \ldots, \otimes_n}$
$\struct {S_2, *_1, *_2, \ldots, *_n}$
$\struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$

Let:

$\phi: \struct {S_1, \otimes_1, \otimes_2, \ldots, \otimes_n} \to \struct {S_2, *_1, *_2, \ldots, *_n}$
$\psi: \struct {S_2, *_1, *_2, \ldots, *_n} \to \struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$

Then the composite of $\phi$ and $\psi$ is also a homomorphism.

## Proof

Let $\psi \circ \phi$ denote the composite of $\phi$ and $\psi$.

Then what we are trying to prove is denoted:

$\paren {\psi \circ \phi}: \struct {S_1, \otimes_1, \otimes_2, \ldots, \otimes_n} \to \struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$ is a homomorphism.

To prove the above is the case, we need to demonstrate that the morphism property is held by each of the operations $\otimes_1, \otimes_2, \ldots, \otimes_n$ under $\psi \circ \phi$.

Let $\otimes_k$ be one of these operations.

We take two elements $x, y \in S_1$, and put them through the following wringer:

 $\ds \map {\paren {\psi \circ \phi} } {x \otimes_k y}$ $=$ $\ds \map \psi {\map \phi {x \otimes_k y} }$ Definition of Composition of Mappings $\ds$ $=$ $\ds \map \psi {\map \phi x *_k \map \phi y}$ Definition of Morphism Property $\ds$ $=$ $\ds \map \psi {\map \phi x} \oplus_k \map \psi {\map \phi y}$ Definition of Morphism Property $\ds$ $=$ $\ds \map {\paren {\psi \circ \phi} } x \oplus_k \map {\paren {\psi \circ \phi} } y$ Definition of Composition of Mappings

Disentangling the confusing and tortuous expressions above, we (eventually) see that this shows that the morphism property is indeed held by $\otimes_k$ under $\psi \circ \phi$.

As this holds for any arbitrary operation $\otimes_k$ in $\struct {S_1, \otimes_1, \otimes_2, \ldots, \otimes_n}$, it follows that it holds for all of them.

Thus $\paren {\psi \circ \phi}: \struct {S_1, \otimes_1, \otimes_2, \ldots, \otimes_n} \to \struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$ is indeed a homomorphism.

$\blacksquare$