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.

$\blacksquare$


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