Direct Image Mapping of Surjection is Surjection/Proof 2

Theorem

Let $f: S \to T$ be a surjection.

$f^\to: \powerset S \to \powerset T$

Let $f: S \to T$ be a surjection.

By definition, $f^\to$ is defined by sending subsets of $S$ to their image under $f$.

That is:

$\forall X \subseteq S: \map {f^\to} X = \set{\map f x: x \in X} \subseteq T$

To prove that $f^\to$ is a surjection, we need to show that every subset of $T$ is the image under $f^\to$ of some subset of $S$.

Let $Y \subseteq T$.

Since $f$ is a surjection, by definition we have that:

$\forall t \in T: \exists s \in S: \map f s = t$

Hence (possibly requiring the Axiom of Choice), we can select for each $y \in Y$ some $s_y \in S$ such that $\map f {s_y} = y$.

Define $X_Y$ to be:

$X_Y := \set {s_y: y \in Y}$

Then:

$\ds \map {f^\to} {X_Y}$

$=$

$\ds \set {\map f {s_y}: s_y \in X_Y}$

$\ds $

$=$

$\ds \set {\map f {s_y}: y \in Y}$

$\ds $

$=$

$\ds \set {y : y \in Y}$

$\ds $

$=$

$\ds Y$

Thus $f^\to: \powerset S \to \powerset T$ is a surjection.

$\blacksquare$

This proof depends on the Axiom of Choice.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.