Surjection if Composite is Surjection

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Theorem

Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be mappings such that $g \circ f$ is a surjection.


Then $g$ is a surjection.


Proof

Let $g \circ f$ be surjective.

Fix $z \in S_3$.

Now find an $x \in S_1: \map {g \circ f} x = z$.

The surjectivity of $g \circ f$ guarantees this can be done.

Now find an $y \in S_2: f \paren x = y$.

$f$ is a mapping and therefore a left-total relation; which guarantees this too can be done.

It follows that:

\(\displaystyle \map g y\) \(=\) \(\displaystyle \map g {\map f x}\)
\(\displaystyle \) \(=\) \(\displaystyle \map {g \circ f} x\) Definition of Composition of Mappings
\(\displaystyle \) \(=\) \(\displaystyle z\) Choice of $x$

$\blacksquare$


Also see


Sources