Embedding Theorem

Theorem
Let:
 * $(1): \quad \struct {T_2, \oplus_2}$ be a submagma of $\struct {S_2, *_2}$
 * $(2): \quad f: \struct {T_1, \oplus_1} \to \struct {T_2, \oplus_2}$ be an isomorphism

then there exists:
 * $(1): \quad$ a magma $\struct {S_1, *_1}$ which algebraically contains $\struct {T_1, \oplus_1}$
 * $(2): \quad g: \struct {S_1, *_1} \to \struct {S_2, *_2}$ where $g$ is an isomorphism which extends $f$.

Proof
There are two cases to consider: when $T_1$ and $S_2$ are disjoint, and when they are not.

Suppose $T_1$ and $S_2$ are disjoint.

Let $S_1$ be the set $T_1 \cup \paren {S_2 \setminus T_2}$.

Then we can define the mapping $h: S_2 \to S_1$ as:


 * $\forall x \in S_2: \map h x = \begin{cases}

x & : x \in S_2 \setminus T_2 \\ \map {f^{-1} } x & : x \in T_2 \end{cases}$

Because $T_1$ and $S_2$ are disjoint, $h$ can be seen to be a bijection.

What $h$ is doing is that it effectively "slots" $T_1$ into the gap in $S_2$ that was taken up by the removed $T_2$.

We are going to show that $\struct {T_1, \oplus_1}$ is embedded in $\struct {S_1, *_1}$.

The operation $*_1$ is defined as the transplant of $*_2$ under $h$.

Thus by the Transplanting Theorem:


 * $\forall x, y \in S_1: x *_1 y = \map h {\map {h^{-1} } x *_2 \map {h^{-1} } y}$

Let $x, y \in T_1$.

Then, from the definition of $h$ above:
 * $\map {h^{-1} } x = \paren {\map {f^{-1} } x}^{-1} = \map f x$

and similarly:
 * $\map {h^{-1} } y = \map f y$

So:

proving that $x \oplus_1 y$ is closed.

Thus:
 * $\struct {T_1, \oplus_1} \subseteq \struct {S_1, *_1}$

that is, $\struct {T_1, \oplus_1}$ is embedded in $\struct {S_1, *_1}$.

Now, let $g = h^{-1}$.

By the definition of $*_1$, $g$ is an isomorphism from $\struct {S_1, *_1}$ onto $\struct {S_2, *_2}$.

Then:

Thus:

So $g$ is an extension of $f$.

We have therefore proved that the embedding theorem holds when $T_1$ and $S_2$ are disjoint.

Next, suppose $T_1$ and $S_2$ are not disjoint.

We can use Exists Bijection to a Disjoint Set to assume the existence of a bijection $k: S_2 \to S_3$ such that $S_3 \cap T = \varnothing$.

Let $*'$ be the transplant of $*$ under $k$.

Let:
 * $T_3 = \set {\map k x: x \in T_2}$

Then:

So $T_3$ is closed under $*'$.

Let $*' {\restriction_{T_3} }$ be the operation on $T_3$ induced by $*'$.

Then $\struct {T_3, *' {\restriction_{T_3} } }$ is embedded in $\struct {S_3, *'}$.

Thus $k \circ f$ is an isomorphism from $\struct {T_1, \oplus_1}$ onto $\struct {T_3, *' {\restriction_{T_3} } }$.

$S_3$ has been constructed so as to be disjoint from $T_1$.

In such a case it has already been shown that the Embedding Theorem holds.

That is:


 * $(1): \quad$ there exists a magma $\struct {S_1, *_1}$ containing $\struct {T_1, \oplus_1}$ algebraically
 * $(2): \quad$ there exists an isomorphism $g_1$ from $\struct {S_1, *_1}$ to $\struct {S_3, \oplus_1}$ extending $k \circ f$

where $\circ$ denotes the composition of $k$ with $f$

Let $g = k^{-1} \circ g_1$.

Let $x \in T$.

Then:

and so $g$ extends $f$.

As $k$ is an isomorphism from $\struct {S_2, *_2}$ onto $\struct {S_3, *'}$, then:


 * $(1): \quad g$ is an isomorphism from $\struct {S_1, *_1}$ onto $\struct {S_2, *_2}$
 * $(2): \quad g$ extends $f$.

Hence the result.