Surjection from Class to Proper Class

Theorem

Let $A$ be a class.

Let $\mathrm P$ be a proper class.

Let $f: A \to \mathrm P$ be a surjection.

Then $A$ is proper.

Aiming for a contradiction, suppose $A$ is not proper.

Then $A$ must be a set.

By the Axiom of Powers, $\powerset A$ is also a set.

Let $g: f \sqbrk A \to \powerset A$ be defined as:

$\map g {\map f a} = f^{-1} \sqbrk {\set {\map f a} }$

It should be noted that:

$\forall a \in A: \map {\paren {g \circ f} } a \ne \O$:

Suppose that $\map g {\map f a} = \map g {\map f b}$.

Let $x \in f^{-1} \sqbrk {\set {\map f a} }$.

Then by the definition of a singleton:

$x \in f^{-1} \sqbrk {\set {\map f a} } \implies \map f x \in \set {\map f a} \implies \map f x = \map f a$

By the same argument:

$x \in f^{-1} \sqbrk {\set {\map f b} } \implies \map f x = \map f b$

$\map f a = \map f b$

It has been shown that $g$ is an injection.

Because $f$ is a surjection, it follows by definition of surjection that:

$f \sqbrk A = \mathrm P$

Thus by contradiction $A$ must be proper.

Hence the result.

$\blacksquare$