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
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.8$