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 $$s_1, s_2 \in S' \Longrightarrow s_1 \circ_k s_2 \in S'$$.

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

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


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


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