Surjection iff Right Cancellable/Necessary Condition/Proof 2
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Theorem
Let $f$ be a surjection.
Then $f$ is right cancellable.
Proof
Let $f: X \to Y$ be surjective.
Then from Surjection iff Right Inverse:
- $\exists g: Y \to X: f \circ g = I_Y$
Suppose $h \circ f = k \circ f$ for two mappings $h: Y \to Z$ and $k: Y \to Z$.
Then:
\(\ds h\) | \(=\) | \(\ds h \circ I_Y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds h \circ \paren {f \circ g}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {h \circ f} \circ g\) | Composition of Mappings is Associative | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k \circ f} \circ g\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds k \circ \paren {f \circ g}\) | Composition of Mappings is Associative | |||||||||||
\(\ds \) | \(=\) | \(\ds k \circ I_Y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds k\) |
Thus $f$ is right cancellable.
So surjectivity implies right cancellability.
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.6$