Problem statement, as written:
Let $f\colon A \rightarrow C$ and $g\colon A \rightarrow B$ be functions. Prove that there exists a function $h\colon B \rightarrow C$ such that $f = h \circ g$ if and only if $\forall x,y \in A$, $g(x) = g(y) \rightarrow f(x) = f(y)$. Prove that $h$ is unique.
The problem is... as written, $h$ isn’t unique. The only way I’ve been able to reconcile this theorem is by adding the premise, “let $g$ be surjective.” Can anyone confirm this? The following problem in the exercise set is here, and, as you can see, includes the phrase “suppose $g$ is bijective” in an analogous place to where “suppose $g$ is surjecrive” would make this problem work. Perhaps a typo or omission?
I can provide a link to my proof upon request- it just seems when my posts get too long you guys run away
If $y \in B$ is not in the range of $g$, it obviously doesn't matter what you take for $h(y)$. So, assuming of course $C$ is not a singleton, what you need for $h$ to be unique is that $g$ is surjective.