Frankly I'm not convinced what I have is technically correct so I was hoping to get some verification.
Suppose we have functions $f:A\rightarrow B$, $r:B\rightarrow C$, and $s:B\rightarrow C$.
Assume that
$r \circ f: A \rightarrow C = s \circ f : A \rightarrow C$
$\iff \forall a \in A, (r \circ f)(a) = (s \circ f)(a)$.
And also that we have a function $f$ that is surjective.
Thus $\forall b \in B, \exists a \in A$ such that $f(a) = b$.
$(r \circ f)(a) = (s \circ f)(a)$
$\iff r(f(a)) = s(f(a)) $
$f(a) = b \in B$
Is this all I need to write the next line?
Therefore, $r(b) = s(b), \forall b \in B$.
Does this actually follow?
$\implies r=s$.
It follows that surjectivity is the necessary and sufficient condition for the hypothesis to be true.