Identity of Cardinal Product is One

Theorem
Let $\mathbf a$ be a cardinal.

Then:
 * $\mathbf 1 \mathbf a = \mathbf a$

where $\mathbf 1 \mathbf a$ denotes the product of the (cardinal) one and $\mathbf a$.

That is, $\mathbf 1$ is the identity element of the product operation on cardinals.

Proof
Let $\mathbf a = \operatorname{Card} \left({A}\right)$ for some set $A$.

From the definition of (cardinal) one, $\mathbf 1$ is the cardinal associated with a singleton set, say, $\left\{{\varnothing}\right\}$.

We have by definition of product of cardinals that $\mathbf 1 \mathbf a$ is the cardinal associated with $\left\{{\varnothing}\right\} \times A$.

Consider the mapping $f: \left\{{\varnothing}\right\} \times A \to A$ defined as:
 * $\forall \left({\varnothing, a}\right) \in \left\{{\varnothing}\right\} \times A: f \left({\varnothing, a}\right) = a$

Let $\left({\varnothing, a_1}\right), \left({\varnothing, a_2}\right) \in \left\{{\varnothing}\right\} \times A$ such that:
 * $f \left({\varnothing, a_1}\right) = f \left({\varnothing, a_2}\right)$

That is:
 * $a_1 = a_2$

It follows from Equality of Ordered Pairs that $\left({\varnothing, a_1}\right) = \left({\varnothing, a_2}\right)$.

That is:
 * $f \left({\varnothing, a_1}\right) = f \left({\varnothing, a_2}\right) \implies \left({\varnothing, a_1}\right) = \left({\varnothing, a_2}\right)$

and so $f$ is injective.

Now let $a \in A$.

By definition:
 * $a = f \left({\varnothing, a}\right)$

and so $f$ is a surjection.

So $f$ is both an injection and a surjection, and so by definition a bijection.

Thus a bijection has been established between $\left\{{\varnothing}\right\} \times A$ and $A$.

It follows by definition that $\left\{{\varnothing}\right\} \times A$ and $A$ are equivalent.

The result follows by definition of cardinal.