Codomain of Internal Direct Isomorphism is Subset Product of Factors

Theorem
Let $\struct {S, \circ}$ be an algebraic structure with $1$ operation.

Let $\struct {A, \circ {\restriction_A} }$ and $\struct {B, \circ {\restriction_B} }$ be closed algebraic substructures of $\struct {S, \circ}$, where $\circ {\restriction_A}$ and $\circ {\restriction_B}$ are the operations induced by the restrictions of $\circ$ to $A$ and $B$ respectively.

Let $\struct {S, \circ}$ be the internal direct product of $A$ and $B$.

Then $S$ is the subset product of $A$ and $B$:
 * $S = A \circ B$

Proof
First we establish that from Set of Finite Subsets under Induced Operation is Closed:
 * $A \times B \subseteq S$

From Condition for Mapping between Structure and Cartesian Product of Substructures to be Bijection:


 * $\phi$ is a bijection


 * for all $s \in S$: there exists a unique $\tuple {a, b} \in A \times B$ such that $a \circ b = s$.
 * for all $s \in S$: there exists a unique $\tuple {a, b} \in A \times B$ such that $a \circ b = s$.

By definition of internal direct product:
 * the mapping $\phi: A \times B \to S$ defined as:


 * $\forall a \in A, b \in B: \map \phi {a, b} = a \circ b$


 * is an isomorphism from the (external) direct product $\struct {A, \circ {\restriction_A} } \times \struct {B, \circ {\restriction_B} }$ onto $\struct {S, \circ}$.

Thus $\phi$ is a bijection.

From Condition for Mapping between Structure and Cartesian Product of Substructures to be Bijection it follows that:
 * for all $s \in S$: there exists a unique $\tuple {a, b} \in A \times B$ such that $a \circ b = s$.

That is:
 * $S = \set {a \circ b: \tuple {a, b} \in A \times B}$

Hence $S = A \circ B$ by definition of subset product.