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 or right cancellable under $\circ_k$, then $\phi \left({a}\right)$ is correspondingly left 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:

and thus left cancellability is demonstrated.


 * And the second is like it, namely this:

and thus right cancellability is demonstrated.