Mapping is Surjection if its Direct Image Mapping is Surjection

Theorem

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

Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$.

Let $f^\to$ be a surjection.

Then $f: S \to T$ is also a surjection.

Proof

Let $y \in S$.

Since $f^\to$ is a surjection:

$\exists U \in \powerset S: \map {f^\to} U = \set y$

From definition of direct image mapping:

$\set y \ne \O \implies U \ne \O$

Let $x \in U$.

Then from definition of direct image mapping:

$\map f x = y$

Since $y$ is arbitrary, $f$ is a surjection.

$\blacksquare$

Also see

• Direct Image Mapping of Surjection is Surjection