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

then:
 * 1) there exists a groupoid $$\left({S_1, *_1}\right)$$ which algebraically contains $$\left({T_1, \oplus_1}\right)$$
 * 2) $$g: \left({S_1, *_1}\right) \to \left({S_2, *_2}\right)$$ 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 $$*'|_{T_3}$$ be the operation on $$T_3$$ induced by $$*'$$.

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

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) there exists a groupoid $$\left({S_1, *_1}\right)$$ containing $\left({T_1, \oplus_1}\right)$ algebraically;
 * 2) 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) $$g$$ is an isomorphism from $$\left({S_1, *_1}\right)$$ onto $$\left({S_2, *}\right)$$;
 * 2) $$g$$ extends $$f$$.