Let $f:A\to B$
Prove the map $f$ is injective iff it has a left inverse.
Starting with $f$ is injective
Let $A$ not be $ \varnothing$
$f(A) = \{b \in B \ | \ b = f(a) \text{ for some } a \in A\}$
If there is a map $g: B \to A$ defined by $g(B) = \{a \in A \ | \ a = g(b)\text{ for some } b \in B\}$ where $b = f(a)$
The composite function $g \circ f(a): a \to f(a) \to a$ returns the identity on $A$ and is thus a left inverse.
$g$ does not exist in this definition if $g(b_i) \neq a_i$ for some $i$ denoting an element of $A,B$, which makes it the nullset
Or
$g(b_i) = a_i = a_{i+x}$ two or more elements of $A$ share an element of $B$ in the mapping of $f: A \to B$ It cannot be true because $a_i \neq a_{i+x}$ implies $f(a_i) \neq f(a_{i+x})$