Homomorphism of External Direct Products

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Theorem

Let:

$\left({S_1 \times S_2, \circ}\right)$ be the external direct product of two algebraic structures $\left({S_1, \circ_1}\right)$ and $\left({S_2, \circ_2}\right)$
$\left({T_1 \times T_2, *}\right)$ be the external direct product of two algebraic structures $\left({T_1, *_1}\right)$ and $\left({T_2, *_2}\right)$
$\phi_1$ be a homomorphism from $\left({S_1, \circ_1}\right)$ onto $\left({T_1, *_1}\right)$
$\phi_2$ be a homomorphism from $\left({S_2, \circ_2}\right)$ onto $\left({T_2, *_2}\right)$.


Then the mapping $\phi_1 \times \phi_2: \left({S_1 \times S_2, \circ}\right) \to \left({T_1 \times T_2, *}\right)$ defined as:

$\left({\phi_1 \times \phi_2}\right) \left({\left({x, y}\right)}\right) = \left({\phi_1 \left({x}\right), \phi_2 \left({y}\right)}\right)$

is a homomorphism from $\left({S_1 \times S_2, \circ}\right)$ to $\left({T_1 \times T_2, *}\right)$.


General Result

Let $n \in \N_{>0}$.

Let:

$\displaystyle \left({\mathcal S_n, \circledcirc_n}\right) := \prod_{k \mathop = 1}^n S_k = \left({S_1, \circ_1}\right) \times \left({S_2, \circ_2}\right) \times \cdots \times \left({S_n, \circ_n}\right)$
$\displaystyle \left({\mathcal T_n, \circledast_n}\right) := \prod_{k \mathop = 1}^n T_k = \left({T_1, \ast_1}\right) \times \left({T_2, \ast_2}\right) \times \cdots \times \left({T_n, \ast_n}\right)$

be external direct products of algebraic structures.


Let $\Phi_n: \left({\mathcal S_n, \circledcirc_n}\right) \to \left({\mathcal T_n, \circledast_n}\right)$ be the mapping defined as:

$\Phi_n: \left({s_1, \ldots, s_n}\right) := \begin{cases} \phi_1 \left({s_1}\right) & : n = 1 \\ \left({\phi_1 \left({s_1}\right), \phi_2 \left({s_2}\right)}\right) & : n = 2 \\ \left({\Phi_n \left({s_1, \ldots, s_{n - 1} }\right), \phi_n \left({s_n}\right)}\right) & : n > 2 \\ \end{cases}$

That is:

$\Phi_n: \left({s_1, \ldots, s_n}\right) := \left({\phi_1 \left({s_1}\right), \phi_2 \left({s_2}\right), \ldots, \phi_n \left({s_n}\right)}\right)$


Let $\phi_k: \left({S_k, \circ_k}\right) \to \left({T_k, \ast_k}\right)$ be a homomorphism for each $k \in \left\{{1, 2, \ldots, n}\right\}$.

Then $\Phi_n$ is a homomorphism from $\left({\mathcal S_n, \circledcirc_n}\right)$ to $\left({\mathcal T_n, \circledast_n}\right)$.


Proof

Let $\left({x_1, x_2}\right), \left({y_1, y_2}\right) \in S_1 \circ S_2$.

Then:

\(\displaystyle \left({\phi_1 \times \phi_2}\right) \left({\left({x_1, x_2}\right) \circ \left({y_1, y_2}\right)}\right)\) \(=\) \(\displaystyle \left({\phi_1 \times \phi_2}\right) \left({\left({x_1 \circ_1 y_1, x_2 \circ_2 y_2}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\phi_1 \left({x_1 \circ_1 y_1}\right), \phi_2 \left({x_2 \circ_2 y_2}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\phi_1 \left({x_1}\right) \ast_1 \phi_1 \left({y_1}\right), \phi_2 \left({x_2}\right) \ast_2 \phi_2 \left({y_2}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\phi_1 \left({x_1}\right), \phi_2 \left({x_2}\right)}\right) \ast \left({\phi_1 \left({y_1}\right), \phi_2 \left({y_2}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\phi_1 \times \phi_2}\right) \left({\left({x_1, x_2}\right)}\right) \ast \left({\phi_1 \times \phi_2}\right) \left({\left({y_1, y_2}\right)}\right)\)

thus demonstrating the morphism property.

$\blacksquare$