Injection iff Left Inverse/Proof 2

Theorem

A mapping $f: S \to T, S \ne \O$ is an injection if and only if:

$\exists g: T \to S: g \circ f = I_S$

where $g$ is a mapping.

That is, if and only if $f$ has a left inverse.

Proof

Take the result Condition for Composite Mapping on Left:

Let $A, B, C$ be sets.

Let $f: A \to B$ and $g: A \to C$ be mappings.

Then:

$\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$

if and only if:

$\exists h: B \to C$ such that $h$ is a mapping and $h \circ f = g$.

Let $C = A = S$, let $B = T$ and let $g = I_S$.

Then the above translates into:

$\exists h: T \to S$ such that $h$ is a mapping and $h \circ f = g$

if and only if:

$\forall x, y \in S: \map f x = \map f y \implies \map {I_S} x = \map {I_S} y$

But as $\map {I_S} x = x$ and $\map {I_S} y = y$ by definition of identity mapping, it follows that:

$\exists h: T \to S$ such that $h$ is a mapping and $h \circ f = g$

if and only if:

$\forall x, y \in S: \map f x = \map f y \implies x = y$

which is our result.

$\blacksquare$

Sources

• 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.4$