Morphism Property Preserves Cancellability
Theorem
Let:
- $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$
be a mapping from one algebraic structure:
- $\struct {S, \circ_1, \circ_2, \ldots, \circ_n}$
to another:
- $\struct {T, *_1, *_2, \ldots, *_n}$
Let $\circ_k$ have the morphism property under $\phi$ for some operation $\circ_k$ in $\struct {S, \circ_1, \circ_2, \ldots, \circ_n}$.
Then if an element $a \in S$ is either left cancellable or right cancellable under $\circ_k$, then $\map \phi a$ is correspondingly left cancellable or right cancellable under $*_k$.
Thus, the morphism property is seen to preserve cancellability.
Proof
First let $S$ be the empty set.
It follows from the definition of an Definition:Morphism Property that $\circ_1, \circ_2, \ldots, \circ_n$ are all empty maps.
It also follows from the definition of an Definition:Morphism Property that $*_1, *_2, \ldots, *_n$ are all empty maps.
By Image of Empty Set is Empty Set, $T$ is also the empty set.
 | This needs considerable tedious hard slog to complete it. In particular: Prove that the morphism property preserves cancellability when $S$ is empty To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Hence, the morphism property is seen to preserve cancellability when $S$ is the empty set.
$\Box$
Now let $S \ne \O$.
We need to demonstrate the following properties:
$(1): \quad$ If $a \in S$ has the property that:
- $\forall x, y \in S: x \circ_k a = y \circ_k a \implies x = y$
then:
- $\forall x, y \in S: \map \phi x *_k \map \phi a = \map \phi y *_k \map \phi a \implies \map \phi x = \map \phi y$
Thus:
| \(\ds x \circ_k a = y \circ_k a\) | \(\implies\) | \(\ds x = y\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map \phi {x \circ_k a} = \map \phi {y \circ_k a}\) | \(\implies\) | \(\ds \map \phi x = \map \phi y\) | Mappings are many-to-one by definition | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map \phi x *_k \map \phi a = \map \phi y *_k \map \phi a\) | \(\implies\) | \(\ds \map \phi x = \map \phi y\) | Definition of Morphism Property |
and thus left cancellability is demonstrated.
$\Box$
$(2): \quad$ If $a \in S$ has the property that:
- $\forall x, y \in S: a \circ_k x = a \circ_k y \implies x = y$
then:
- $\forall x, y \in S: \map \phi a *_k \map \phi x = \map \phi a *_k \map \phi y \implies \map \phi x = \map \phi y$
Thus:
| \(\ds a \circ_k x = a \circ_k y\) | \(\implies\) | \(\ds x = y\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map \phi {a \circ_k x} = \map \phi {a \circ_k y}\) | \(\implies\) | \(\ds \map \phi x = \map \phi y\) | Mappings are many-to-one by definition | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map \phi a *_k \map \phi x = \map \phi a *_k \map \phi y\) | \(\implies\) | \(\ds \map \phi x = \map \phi y\) | Definition of Morphism Property |
and thus right cancellability is demonstrated.
$\blacksquare$