Order of Subset Product with Singleton/Proof 2

Proof
Let $\order Y = k$.

We define the mapping $\phi: Y \to X \circ Y$ such that:
 * $\forall y \in Y: \map \phi y = x \circ y$

First we show that $\phi$ is injective.

Let $y_1, y_2 \in Y$.

Let $\map \phi {y_1} = \map \phi {y_2}$.

Hence $\phi$ is injective.

The fact that $\phi$ is surjective follows from the definition of $X \circ Y$.

Every element of $X \circ Y$ is of the form $x \circ y$ for some $y \in Y$.

Thus $\phi$ is surjective.

The other half of the result follows identically, by defining a similar function for $Y \circ X$.