Morphism Property Preserves Cancellability

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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.



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.

$\blacksquare$