Direct Image Mapping of Surjection is Surjection/Proof 3
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Theorem
Let $f: S \to T$ be a surjection.
Then the direct image mapping of $f$:
- $f^\to: \powerset S \to \powerset T$
is a surjection.
Proof
Let $f^\gets$ be the inverse image mapping of $f$.
Let $Y \in \powerset T$.
Let $X = \map {f^\gets} Y$.
By Subset equals Image of Preimage iff Mapping is Surjection:
- $\map {f^\to} X = Y$
As such an $X$ exists for each $Y \in \powerset S$, $f^\to$ is surjective.
$\blacksquare$