# Image of Preimage of Subset under Surjection equals Subset

## Theorem

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

Then:

$\forall B \subseteq T: B = \paren {f \circ f^{-1} } \sqbrk B$

where:

$f \sqbrk B$ denotes the image of $B$ under $f$
$f^{-1}$ denotes the inverse of $f$
$f \circ f^{-1}$ denotes composition of $f$ and $f^{-1}$.

## Proof

Let $g$ be a surjection.

Let $B \subseteq T$.

Let $b \in B$.

Then:

 $\ds \exists a \in S: \,$ $\ds b$ $=$ $\ds \map f a$ Definition of Surjection $\ds \leadsto \ \$ $\ds a$ $\in$ $\ds f^{-1} \sqbrk B$ Definition of Preimage of Subset under Mapping $\ds \leadsto \ \$ $\ds b$ $\in$ $\ds f \sqbrk {f^{-1} \sqbrk B}$ Definition of Image of Subset under Mapping $\ds \leadsto \ \$ $\ds B$ $\subseteq$ $\ds f \sqbrk {f^{-1} \sqbrk B}$ Definition of Subset $\ds \leadsto \ \$ $\ds B$ $\subseteq$ $\ds \paren {f \circ f^{-1} } \sqbrk B$ Definition of Composition of Mappings

From Subset of Codomain is Superset of Image of Preimage, we already have that:

$\paren {f \circ f^{-1} } \sqbrk B \subseteq B$

So:

$B \subseteq \paren {f \circ f^{-1} } \sqbrk B \subseteq B$

and by definition of set equality:

$B = \paren {f \circ f^{-1} } \sqbrk B$

$\blacksquare$