# 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 $*_k$:

 $\ds t_1, t_2$ $\in$ $\ds \phi \sqbrk {S'}$ $\ds \leadsto \ \$ $\ds \exists s_1 \in S': \,$ $\ds t_1$ $=$ $\ds \map \phi {s_1}$ Definition of Image of Subset under Mapping $\ds \land \ \$ $\ds \exists s_2 \in S': \,$ $\ds t_2$ $=$ $\ds \map \phi {s_2}$ $\ds \leadsto \ \$ $\ds t_1 *_k t_2$ $=$ $\ds \map \phi {s_1} *_k \map \phi {s_2}$ $\ds$ $=$ $\ds \map \phi {s_1 \circ_k s_2}$ Definition of Morphism Property $\ds \leadsto \ \$ $\ds t_1 *_k t_2$ $\in$ $\ds \phi \sqbrk {S'}$ $S'$ is closed under $\circ$

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

 $\ds s_1, s_2$ $\in$ $\ds \phi^{-1} \sqbrk {T'}$ $\ds \leadsto \ \$ $\ds \map \phi {s_1}, \map \phi {s_2}$ $\in$ $\ds T'$ Definition of Inverse Mapping $\ds \map \phi {s_1} *_k \map \phi {s_2}$ $\in$ $\ds T'$ $\ds \leadsto \ \$ $\ds \map \phi {s_1 \circ_k s_2}$ $\in$ $\ds T'$ Definition of Morphism Property $\ds \leadsto \ \$ $\ds s_1 \circ_k s_2$ $\in$ $\ds \phi^{-1} \sqbrk {T'}$ Definition of Inverse Mapping

$\blacksquare$