Morphism Property Preserves Closure

Theorem

Let $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({T, *_1, *_2, \ldots, *_n}\right)$ be a mapping from one algebraic structure $\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$ to another $\left({T, *_1, *_2, \ldots, *_n}\right)$.

Let $\circ_k$ have the morphism property under $\phi$ for some operation $\circ_k$ in $\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$.

Then the following properties hold:

• If $S\,' \subseteq S$ is closed under $\circ_k$, then $\phi \left({S\,'}\right)$ is closed under $*_k$.
• If $T\,' \subseteq T$ is closed under $*_k$, then $\phi^{-1} \left({T\,'}\right)$ is closed under $\circ_k$.

Proof

Suppose that $\circ_k$ has the morphism property under $\phi$.

Suppose that $S\,' \subseteq S$ is closed under $\circ_k$.

Thus $s_1, s_2 \in S\,' \implies s_1 \circ_k s_2 \in S\,'$.

Similarly, suppose that $T\,' \subseteq T$ is closed under $*_k$.

Thus $t_1, t_2 \in T\,' \implies t_1 *_k t_2 \in T\,'$.

• First we prove that $\phi \left({S\,'}\right)$ is closed under $*_k$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle t_1, t_2\)	\(\in\)	\(\displaystyle \phi \left({S\,'}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \exists s_1 \in S\,': t_1\)	\(=\)	\(\displaystyle \phi \left({s_1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Image of Subset
	\(\displaystyle \)	\(\displaystyle \land\)	\(\displaystyle \)	\(\displaystyle \exists s_2 \in S\,': t_2\)	\(=\)	\(\displaystyle \phi \left({s_2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle t_1 *_k t_2\)	\(=\)	\(\displaystyle \phi \left({s_1}\right) *_n \phi \left({s_2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi \left({s_1 \circ_k s_2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of morphism property
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle t_1 *_k t_2\)	\(\in\)	\(\displaystyle \phi \left({S\,'}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $S\,'$ is closed under $\circ$

• Then we prove that $\phi^{-1} \left({T\,'}\right)$ is closed under $\circ_k$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle s_1, s_2\)	\(\in\)	\(\displaystyle \phi^{-1} \left({T\,'}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \phi \left({s_1}\right), \phi \left({s_2}\right)\)	\(\in\)	\(\displaystyle T\,'\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of inverse mapping
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \phi \left({s_1}\right) *_k \phi \left({s_2}\right)\)	\(\in\)	\(\displaystyle T\,'\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \phi \left({s_1 \circ_k s_2}\right)\)	\(\in\)	\(\displaystyle T\,'\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of morphism property
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle s_1 \circ_k s_2\)	\(\in\)	\(\displaystyle \phi^{-1} \left({T\,'}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of inverse mapping

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 12$: Theorem $12.1$