Embedding Theorem
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Context
The following is a frequently occurring circumstance in the field of abstract algebra.
- We have a magma $\left({T_1, \circ}\right)$.
- $\left({T_1, \circ}\right)$ is isomorphic to another magma $\left({T_2, *}\right)$.
- $\left({T_2, *}\right)$ is embedded in a magma $\left({S_2, *}\right)$.
- We want to embed $\left({T_1, \circ}\right)$ in its own magma $\left({S_1, \circ}\right)$ 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 submagma 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 magma $\left({S_1, *_1}\right)$ 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 $f$.
Add the corollary that extends this theorem to structures with two operations.
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Proof
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 $T_1 \cup \left({S_2 \setminus T_2}\right)$.
Then we can define the mapping $h: S_2 \to S_1$ as:
- $\forall x \in S_2: h \left({x}\right) = \begin{cases} x & : x \in S_2 \setminus T_2 \\ f^{-1} \left({x}\right) & : x \in T_2 \end{cases}$
Because $T_1$ and $S_2$ are disjoint, $h$ can be seen to be a bijection.
What $h$ is doing is that it 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 $\left({T_1, \oplus_1}\right)$ is embedded in $\left({S_1, *_1}\right)$.
The operation $*_1$ is defined as the transplant of $*_2$ under $h$.
Thus by the Transplanting Theorem:
- $\forall x, y \in S_1: x *_1 y = h \left({h^{-1} \left({x}\right) *_2 h^{-1} \left({y}\right)}\right)$
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:
| \(\displaystyle x *_1 y\) | \(=\) | \(\displaystyle h \left({f \left({x}\right) *_2 f \left({y}\right)}\right)\) | Transplanting Theorem | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle h \left({f \left({x}\right) \oplus_2 f \left({y}\right)}\right)\) | as $*_2$ is embedded in $\oplus_2$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle h \left({f \left({x \oplus_1 y}\right)}\right)\) | Morphism Property of $\oplus_1$ under $f$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle f^{-1} \left({f \left({x \oplus_1 y}\right)}\right)\) | Definition of $h$ on elements of $T_1$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle x \oplus_1 y\) | Inverse of Inverse of Bijection |
proving that $x \oplus_1 y$ is closed.
Thus:
- $\left({T_1, \oplus_1}\right) \subseteq \left({S_1, *_1}\right)$
that is, $\left({T_1, \oplus_1}\right)$ is embedded in $\left({S_1, *_1}\right)$.
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:
| \(\displaystyle x \in T_1\) | \(\implies\) | \(\displaystyle f \left({x}\right) \in T_2\) | by definition of $f$ | ||||||||||
| \(\displaystyle \) | \(\implies\) | \(\displaystyle h \left({f \left({x}\right)}\right) = f^{-1} \left({f \left({x}\right)}\right) = x\) | by definition of $h$ |
Thus:
| \(\displaystyle g \left({x}\right)\) | \(=\) | \(\displaystyle g \left({h \left({f \left({x}\right)}\right)}\right)\) | from above | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle h^{-1} \left({h \left({f \left({x}\right)}\right)}\right)\) | by definition of $g$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle f \left({x}\right)\) | Inverse of Inverse of Bijection |
So $g$ is an extension of $f$.
We have therefore proved that the embedding theorem holds when $T_1$ and $S_2$ are disjoint.
$\Box$
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:
- $T_3 = \left\{{k \left({x}\right): x \in T_2}\right\}$
Then:
| \(\displaystyle x, y \in T_3\) | \(\implies\) | \(\displaystyle k^{-1} \left({x}\right), k^{-1} \left({y}\right) \in T_2\) | by definition of $k$ | ||||||||||
| \(\displaystyle \) | \(\implies\) | \(\displaystyle k^{-1} \left({x}\right) * k^{-1} \left({y}\right) \in T_2\) | by definition of $*$ in $T_2$ | ||||||||||
| \(\displaystyle \) | \(\implies\) | \(\displaystyle x *' y = k \left({k^{-1} \left({x}\right) * k^{-1} \left({y}\right)}\right) \in T_3\) | Transplanting Theorem |
So $T_3$ is closed under $*'$.
Let $*' {\restriction_{T_3} }$ be the operation on $T_3$ induced by $*'$.
Then $\left({T_3, *' {\restriction_{T_3} } }\right)$ is embedded in $\left({S_3, *'}\right)$.
Thus $k \circ f$ is an isomorphism from $\left({T_1, \oplus_1}\right)$ onto $\left({T_3, *' {\restriction_{T_3} } }\right)$.
$S_3$ 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 $\left({S_1, *_1}\right)$ 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 $g = k^{-1} \bullet g_1$.
Let $x \in T$.
Then:
| \(\displaystyle g \left({x}\right)\) | \(=\) | \(\displaystyle k^{-1} \left({g_1 \left({x}\right)}\right)\) | definition of $g$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle k^{-1} \left({k \left({f \left({x}\right)}\right)}\right)\) | as $g_1$ extends $k \bullet f$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle f \left({x}\right)\) | Inverse of Inverse of Bijection |
and so $g$ 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 $f$.
Hence the result.
$\blacksquare$
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Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): $\S 8$: Theorem $8.1$
