Subset equals Image of Preimage implies Surjection/Proof 1

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Theorem

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

Let:

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

where $f \sqbrk B$ denotes the image of $B$ under $f$.

Then $f$ is a surjection.


Proof

Let $f$ be such that:

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

In particular, it holds for $T$ itself.

Hence:

\(\ds T\) \(=\) \(\ds \paren {f \circ f^{-1} } \sqbrk T\)
\(\ds T\) \(=\) \(\ds f \sqbrk {f^{-1} \sqbrk T}\) Definition of Composition of Mappings
\(\ds \) \(\subseteq\) \(\ds f \sqbrk S\) Image of Subset under Mapping is Subset of Image
\(\ds \) \(=\) \(\ds \Img f\) Definition of Image of Mapping
\(\ds \) \(\subseteq\) \(\ds T\) Image is Subset of Codomain: Corollary $1$

So:

$T \subseteq \Img f \subseteq T$

and so by definition of set equality:

$\Img f = T$

So, by definition, $f$ is a surjection.

$\blacksquare$


Sources