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
Suppose $S$ is 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.

Suppose $S$ is non-empty.

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:
 * $\forall x, y \in S: \map \phi x *_k \map \phi a = \map \phi y *_k \map \phi a \implies \map \phi x = \map \phi y$

and thus left cancellability is demonstrated.

$(2): \quad$ 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:
 * $\forall x, y \in S: \map \phi a *_k \map \phi x = \map \phi a *_k \map \phi y \implies \map \phi x = \map \phi y$

and thus right cancellability is demonstrated.