Embedding Theorem

Context
The following is a frequently occurring circumstance in the field of abstract algebra.


 * We have a groupoid $\left({T_1, \circ}\right)$.
 * $\left({T_1, \circ}\right)$ is isomorphic to another groupoid $\left({T_2, *}\right)$.
 * $\left({T_2, *}\right)$ is embedded in a groupoid $\left({S_2, *}\right)$.
 * We want to embed $\left({T_1, \circ}\right)$ in its own groupoid $\left({S_1, \circ}\right)$ such that $\left({S_1, \circ}\right) \cong \left({S_2, *}\right)$.

This can always be done, as this theorem shows.

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

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

Proof
Note: In order to reduce notational confusion, the composition of two mappings $f$ and $g$ is denoted $f \bullet g$ instead of the more usual $f \circ g$.

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 $T_1 \cup \left({S_2 - T_2}\right)$.

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

$\forall x \in S_2: h \left({x}\right) = \begin{cases} x & : x \in S_2 - T_2 \\ f^{-1} \left({x}\right) & : x \in T_2 \end{cases}$

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

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

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

Now, we define $*_1$ to be the transplant of $*_2$ under $h$.

By the Transplanting Theorem:

$\forall x, y \in S_1: x *_1 y = h \left({h^{-1} \left({x}\right) *_2 h^{-1} \left({y}\right)}\right)$

Now let $x, y \in T_1$. Then, from the definition of $h$ above:

$h^{-1} \left({x}\right) = \left({f^{-1} \left({x}\right)}\right)^{-1} = f \left({x}\right)$

and similarly $h^{-1} \left({y}\right) = f \left({y}\right)$.

So:

... proving that $x \oplus_1 y$ is closed and therefore $\left({T_1, \oplus_1}\right) \subseteq \left({S_1, *_1}\right)$, i.e. is embedded in it.

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

By the definition of $*_1$, $g$ is an isomorphism from $\left({S_1, *_1}\right)$ onto $\left({S_2, *_2}\right)$. 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 $S_3 \cap T = \varnothing$.

We let $*'$ be the transplant of $*$ under $k$, and let:

$T_3 = \left\{{k \left({x}\right): x \in T_2}\right\}$

Then:

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

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

Then $\left({T_3, *' \restriction_{T_3}}\right)$ is embedded in $\left({S_3, *'}\right)$ and $k \circ f$ is an isomorphism from $\left({T_1, \oplus_1}\right)$ onto $\left({T_3, *' \restriction_{T_3}}\right)$.

Now we have just constructed $S_3$ so it is disjoint from $T_1$, and we have just shown that the embedding theorem holds.

That is:


 * $(1): \quad$ there exists a groupoid $\left({S_1, *_1}\right)$ containing $\left({T_1, \oplus_1}\right)$ algebraically;
 * $(2): \quad$ there exists an isomorphism $g_1$ from $\left({S_1, *_1}\right)$ to $\left({S_3, \oplus_1}\right)$ extending $k \bullet f$.

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

Now if $x \in T$:

... therefore $g$ extends $f$.

As $k$ is an isomorphism from $\left({S_2, *}\right)$ onto $\left({S_3, *'}\right)$, then:


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