Image of Preimage of Subset under Surjection equals Subset

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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:

\(\displaystyle \exists a \in S: \ \ \) \(\displaystyle b\) \(=\) \(\displaystyle \map f a\) Definition of Surjection
\(\displaystyle \leadsto \ \ \) \(\displaystyle a\) \(\in\) \(\displaystyle f^{-1} \sqbrk B\) Definition of Preimage of Subset under Mapping
\(\displaystyle \leadsto \ \ \) \(\displaystyle b\) \(\in\) \(\displaystyle f \sqbrk {f^{-1} \sqbrk B}\) Definition of Image of Subset under Mapping
\(\displaystyle \leadsto \ \ \) \(\displaystyle B\) \(\subseteq\) \(\displaystyle f \sqbrk {f^{-1} \sqbrk B}\) Definition of Subset
\(\displaystyle \leadsto \ \ \) \(\displaystyle B\) \(\subseteq\) \(\displaystyle \paren {f \circ f^{-1} } \sqbrk B\) 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:

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

$\blacksquare$


Also see


Sources