Isomorphism Preserves Left Cancellability

Theorem
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.

Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.

Then:
 * $a \in S$ is left cancellable in $\struct {S, \circ}$


 * $\map \phi a \in T$ is left cancellable in $\struct {T, *}$.
 * $\map \phi a \in T$ is left cancellable in $\struct {T, *}$.

Proof
Let $a$ be left cancellable in $\struct {S, \circ}$.

Let $x \in S$ and $y \in S$ be arbitrary.

Then:

That is, $\map \phi a$ is left cancellable in $\struct {T, *}$.

As $\phi$ is an isomorphism, then so is $\phi^{-1}$.

So the same proof works in reverse in exactly the same way.