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 $\phi$.

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

1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.1$
1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.2$