does this imply that if $f:A→B$ is injective that any mapping
$g:{\rm Im}(f)→A$ is bijective?
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.
I don't understand why we can even use $f^{−1}$ on an element.
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.
How would you go about showing that $f:A→B$ is injective $\implies
f:A→{\rm Im}(f)$ is bijective?
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.
First, the composition is only possible when the codomain of $f$ is a subset of the domain of $g$. This means that $f: A \rightarrow D$ and $g: B \rightarrow C$ with $D\subseteq B$.
Now, to prove part (a), use the definition of injective; $f$ is injective if $f \left( x_1 \right) = f \left( x_2 \right) \Rightarrow x_1 = x_2$.
To use it, assume $\left( g \circ f \right) \left( x_1 \right) = \left( g \circ f \right) \left( x_2 \right)$. This means that $g \left( f \left( x_1 \right) \right) = g \left( f \left( x_2 \right) \right)$. Since $g$ is injective this implies $f \left( x_1 \right) = f \left( x_2 \right)$ and $f$ being injective further implies $x_1 = x_2$. Hence, $g \circ f$ is injective.
For the second part, if you take any element $c \in C$, then the surjectivity of $g \circ f$ implies the existence of $a \in A$ such that $\left( g \circ f \right) \left( a \right) = g \left( f \left( a \right) \right)=c$. Since the codomain of $f$ is the same as domain of $g$ (so that the composition is possible), $f \left( a \right) = b \in B$. Hence, $\forall c \in C$, $\exists b \in B$ such that $g \left( b \right) = c$. In particular, this $b$ is given by $f \left( a \right)$, where $\left( g \circ f \right) \left( a \right) = c$.
Best Answer
Indeed, you need to define $g$ in order to show that a left inverse exists. What lacks in your definition of $g$ is how an element of $B \diagdown f(A)$ is mapped to $A$. Of course, $B \diagdown f(A)$ could be the empty set, but for the general case, we don't know.
Since $A \neq \emptyset$, let $k \in A$. Since $f$ is injective, it follows that $h: f(A) \to A, b \mapsto f^{-1}(b)$ is bijective. Then we define $g(y)=\begin{cases} k & y \in B\diagdown f(A) \\ h(y) & y \in f(A)\end{cases}$
Now we get for all $a\in A$ that $g(f(a))=a$, so $g \circ f = id_{A}$.