Mapping Induced on Power Set by Surjection

Theorem

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

Then the mapping induced by $g$ on $\mathcal P \left({S}\right)$:

$f_g: \mathcal P \left({S}\right) \to \mathcal P \left({T}\right)$

is a surjection.

Proof 1

Suppose $g: S \to T$ is a surjection.

Then $\forall y \in T: \exists x \in S: g \left({x}\right) = y$.

From the Quotient Theorem for Surjections, there is one and only one bijection $r: S / \mathcal R_g \to T$ such that $r \circ q_{\mathcal R_g} = g$.

Each element of $S / \mathcal R_g$ is a subset of $S$ and therefore an element of $\mathcal P \left({S}\right)$.

Thus:

$\forall X_1, X_2 \in \mathcal P \left({S}\right): r \left({X_1}\right) = r \left({X_2}\right) \implies X_1 = X_2$.

Because $g$ is a surjection, every $y \in T$ is mapped to by exactly one element of the partition of $S$ defined by $\mathcal R_g$.

Let $T = \left\{{y_1, y_2, \ldots}\right\}$.

Let the partition defined by $\mathcal R_g$ be $\bigcup \left({X_1, X_2, \ldots}\right)$ where $r \left({X_n}\right) = y_n$.

Let $Y_r \in \mathcal P \left({T}\right)$, such that $Y_r = \left\{{y_{r_1}, y_{r_2}, \ldots}\right\}$.

Then $f_g \left({X_r}\right) = Y_r$, where $X_r = \bigcup \left({X_{r_1}, X_{r_2}, \ldots}\right)$.

As $\left\{{X_1, X_2, \ldots}\right\}$ is a partition of $S$, $\forall Y_r \in \mathcal P \left({T}\right): X_r$ is unique.

Thus $f_g: \mathcal P \left({S}\right) \to \mathcal P \left({T}\right)$ is a surjection.

$\blacksquare$

Proof 2

Suppose $g: S \to T$ is a surjection.

By definition, $f_g$ is defined by sending subsets of $S$ to their image under $g$.

That is, for all $X\subseteq S$, $f_g(X) = \{g(x):x\in X\}\subseteq T$

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

Let $Y\subseteq T$.

Since $g$ is a surjection, by definition we have that $\forall t \in T: \exists s \in S: g \left({s}\right) = t$.

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

Define $X_Y$ to be $\{s_y : y\in Y\}$.

Then $f_g (X_Y) = \{g(s_y) : s_y \in X_Y \} = \{g(s_y) : y\in Y \} = \{y : y\in Y \} = Y$.

Thus $f_g: \mathcal P \left({S}\right) \to \mathcal P \left({T}\right)$ is a surjection.

$\blacksquare$

Axiom of Choice

This proof depends on the Axiom of Choice.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered 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.