Injection has Surjective Left Inverse Mapping/Proof 1
Theorem
Let $S$ and $T$ be sets such that $S \ne \O$.
Let $f: S \to T$ be a injection.
Then there exists a surjection $g: T \to S$ such that:
- $g \circ f = I_S$
Proof
Since $S$ is non-empty, we can choose an element $x \in S$.
Since $f$ is an injection, for each $t \in \Img f$ there exists a unique $s \in S$ such that:
- $\map f s = t$
Thus by Law of Excluded Middle there exists a well-defined mapping $T \to S$ given by:
- $\map g t = \begin {cases} s & : \paren {t \in \Img f} \land \paren {\map f s = t} \\ x & : t \notin \Img f \end {cases}$
By construction, for any given $s \in S$, the element $\map f s$ maps to $s$ under $g$.
Therefore $g: T \to S$ is a surjection.
$\blacksquare$
Law of the Excluded Middle
This theorem depends on the Law of the Excluded Middle.
This is one of the axioms of logic that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.
However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom.
This in turn invalidates this theorem from an intuitionistic perspective.