# Morphism Property Preserves Cancellability

## Theorem

Let:

$\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({T, *_1, *_2, \ldots, *_n}\right)$

be a mapping from one algebraic structure:

$\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$

to another:

$\left({T, *_1, *_2, \ldots, *_n}\right)$

Let $\circ_k$ have the morphism property under $\phi$ for some operation $\circ_k$ in $\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$.

Then if an element $a \in S$ is either left cancellable or right cancellable under $\circ_k$, then $\phi \left({a}\right)$ is correspondingly left cancellable or right cancellable under $*_k$.

Thus, the morphism property is seen to preserve cancellability.

## Proof

We need to demonstrate the following properties:

• 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:

$\forall x, y \in S: \phi \left({x}\right) *_k \phi \left({a}\right) = \phi \left({y}\right) *_k \phi \left({a}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)$

• 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:

$\forall x, y \in S: \phi \left({a}\right) *_k \phi \left({x}\right) = \phi \left({a}\right) *_k \phi \left({y}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)$

• To show the first of the properties above:
 $\ds$  $\ds \left({x \circ_k a = y \circ_k a \implies x = y}\right)$ $\ds$ $\implies$ $\ds \left({\phi \left({x \circ_k a}\right) = \phi \left({y \circ_k a}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)}\right)$ Mappings are many-to-one by definition $\ds$ $\implies$ $\ds \left({\phi \left({x}\right) *_k \phi \left({a}\right) = \phi \left({y}\right) *_k \phi \left({a}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)}\right)$ Morphism Property

and thus left cancellability is demonstrated.

• And the second is like it, namely this:
 $\ds$  $\ds \left({a \circ_k x = a \circ_k y \implies x = y}\right)$ $\ds$ $\implies$ $\ds \left({\phi \left({a \circ_k x}\right) = \phi \left({a \circ_k y}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)}\right)$ Mappings are many-to-one by definition $\ds$ $\implies$ $\ds \left({\phi \left({a}\right) *_k \phi \left({x}\right) = \phi \left({a}\right) *_k \phi \left({y}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)}\right)$ Morphism Property

and thus right cancellability is demonstrated.

$\blacksquare$