Image of Preimage of Subset under Surjection equals Subset

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Theorem

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


Then:

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

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:

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

$\blacksquare$


Also see


Sources

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