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:

$$\forall x, y \in S: x \circ_k a = y \circ_k a \Longrightarrow x = y$$

then:

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


 * If $$a \in S$$ has the property that:

$$\forall x, y \in S: a \circ_k x = a \circ_k y \Longrightarrow x = y$$

then:

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


 * To show the first of the properties above:

and thus left cancellability is demonstrated.


 * The second is shown similarly:

and thus right cancellability is demonstrated.