Surjection from Class to Proper Class

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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.


Proof

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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:

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

Hhence by the properties of equality:

$\map f a = \map f b$


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

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

$f \sqbrk A = \mathrm P$


But this contradicts Injection from Proper Class to Class.

Thus by contradiction $A$ must be proper.


Hence the result.

$\blacksquare$