Morphism Property Preserves Closure

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

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

Then the following properties hold:


 * If $S' \subseteq S$ is closed under $\circ_k$, then $\phi \sqbrk {S'}$ is closed under $*_k$
 * If $T' \subseteq T$ is closed under $*_k$, then $\phi^{-1} \sqbrk {T'}$ is closed under $\circ_k$

where $\phi \sqbrk {S'}$ denotes the image of $S'$.

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

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

Thus, for non-empty $S'$:
 * $s_1, s_2 \in S' \implies s_1 \circ_k s_2 \in S'$

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

Thus, non-empty $T'$:
 * $t_1, t_2 \in T' \implies t_1 *_k t_2 \in T'$

First we prove that $\phi \sqbrk {S'}$ is closed under $*_k$:

Then we prove that $\phi^{-1} \sqbrk {T'}$ is closed under $\circ_k$: