Morphism Property Preserves Closure

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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 $\phi$.


This theorem requires a proof.
In particular: It remains to be shown that the Theorem is true where S is the empty set
The empty case seems considered, doesn't it?
Please feel free to take this debate to the talk page. I also have no idea why we need explicitly to consider the empty set. Suspect AI.
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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

$\blacksquare$


Sources

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