I've already done a lot of searching (in particular: https://www.proofwiki.org/wiki/Injection_iff_Left_Inverse) to try to prove this statement:
$f: A \to B$ is injective if and only if it has a left inverse. [I'm going to also assume $A$ and $B$ are nonempty.]
My attempt so far:
$\Leftarrow$ Suppose $f$ has a left inverse and $f(a_1) = f(a_2)$ for
$a_1, a_2 \in A$. Then there is a function $g: B \to A$ such that
$g\left(f(a_1)\right) = a_1$ and $g\left(f(a_2)\right) = a_2$. Since
$g$ is well-defined, it follows that $g\left(f(a_1)\right) =
g\left(f(a_2)\right)$, and hence $a_1 = a_2$.$\Rightarrow$Now
suppose $f$ is injective. Then $f(a_1) = f(a_2) \implies a_1 = a_2$.
Furthermore, since $A$ is nonempty, there is an element $x \in A$.
Define $g: B \to A$ by \begin{equation*} g(b) = \begin{cases}
f^{-1}(b), & b \in \text{Im}(f) \\ x, & b \in A \setminus
\text{Im}(f)\text{.} \end{cases} \end{equation*} This mapping is
well-defined, since if $f$ is injective, $f^{-1}(b)$ is unique.
My main question: does this imply that if $f: A \to B$ is injective that any mapping $g: \text{Im}(f) \to A$ is bijective? Otherwise, I don't understand why we can even use $f^{-1}$ on an element. [Please read the link above for more details – in proof 1.]
Best Answer
No. It only means that $f: A \to f(A) = {\rm Im}(f)$ is bijective. So you can consider the inverse, but with its domain restricted to the image of the initial function. So you can have more than one left inverse. The big theorem is that if exists both the left and right inverses, then they're equal. Any injective function is a bijection between its domain and its image. It is not true that any mapping $g: {\rm Im}(f) \to A$ is bijective.
In general you can't. Note that we only use $f^{-1}$ where it is well-defined, that is: in the image of $f$. It is well defined because $f$ is injective.
Let's do all the details. Formally, two functions are equal if and only if all the domains, codomains, and rules of association are equals. Let $f:A \to B$ be an injective function. Consider $\bar{f}:A \to {\rm Im}(f)$ be defined by $\bar{f}(x) := f(x)$, for all $x \in A$. I'm going to prove that $\bar{f}$ is bijective.
For injectivity, take $x,y \in A$ such that $\bar{f}(x) = \bar{f}(y)$. So: $$\bar{f}(x) = \bar{f}(y) \implies f(x) = f(y) \implies x = y,$$ this last step being because $f$ is assumed injective.
For surjectivity, let $y \in {\rm Im}(f)$. By definition of image, exists $x \in A$ such that $f(x) = y$. But that means that $\bar{f}(x) = y$, so $\bar{f}$ is surjective.