In category theory, the expression $f\circ g=h$ ($\circ$ being the binary composition "function" on the class of morphisms) suggests that the morphism $h$ is unique when it clearly needn't be. Do we not have to define an equivalence relation on the set of morphisms using the $hom$ class?
Do you have some sort of counterexample or heuristic reasoning that leads you to believe that $h$ is not unique? Given a concrete category, the binary composition of morphisms becomes the binary composition of functions. Since a function takes one input and relates it to exactly one output, applying $g$ to the input $x$ gives the unique output $g(x)$, and the composition $f \circ g$ applied to $x$ gives $f(g(x))$, which is the unique output of $f$ applied to $g(x)$. Since this is the case for all objects $x$ in the concrete category in question, $f \circ g$ has been uniquely determined insofar as I can see.
The common definition of a category demands that for every morphism $f$ there are uniquely determined domain a codomain objects. So, strictly speaking for every objects $A,B,C,D$ in a category if $Hom(A,B)\cap Hom(C,D)\ne \emptyset$ then $A=C$ and $B=D$.
Most definitions of (set enriched) categories will include either the explicit domain/codomain (large) functions or the pair-wise disjoint condition on the hom-sets as axioms (each can be proved from the other of course).
Side remark: Interestingly, in the definition of enriched categories these conditions do not appear explicitly.