# 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. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{ProofWanted}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

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 $12.2$