Surjection iff Right Cancellable/Sufficient Condition/Proof 1
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Theorem
Let $f$ be a mapping which is right cancellable.
Then $f$ is a surjection.
Proof
Suppose $f$ is a mapping which is not surjective.
Then:
- $\exists y_1 \in Y: \neg \exists x \in X: \map f x = y_1$
Let $Z = \set {a, b}$.
Let $h_1$ and $h_2$ be defined as follows.
- $\map {h_1} y = a: y \in Y$
- $\map {h_2} y = \begin {cases}
a & : y \ne y_1 \\ b & : y = y_1 \end {cases}$
Thus we have $h_1 \ne h_2$ such that $h_1 \circ f = h_2 \circ f$.
Therefore $f$ is not right cancellable.
It follows from the Rule of Transposition that if $f$ is right cancellable, then $f$ must be surjective.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $4$: Mappings: Exercise $16$