Image of Preimage of Subset under Surjection equals Subset

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

Then:
 * $\forall B \subseteq T: B = \left({f \circ f^{-1} }\right) \sqbrk B$

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:

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

Also see

 * Subset equals Image of Preimage implies Surjection
 * Subset equals Image of Preimage iff Mapping is Surjection