Is this a valid uniqueness up to isomorphism?

Question

I've shown around for a couple of times asking more or less same "uniqueness up to isomorphism"-related questions. Unfortunately, have to ask yet another one.

Say, there are distinguishable $A, B, C$ sets along with $f : A \mapsto B$ and $g : B \mapsto C$. Those conditions are sufficient to build up unique $h : A \mapsto C, h = g \circ f$.

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1) May I state that $A$ and $B$ are "unique up to isomorphism" since there is a way to map them both to the same $C$?

2) May I state that maps $g$ and $h$ are "unique up to isomorphism" because of the fact that both resulting to the same $C$?

3) Finally, does the term "isomorphism" always equal to the "bijection"? Could any of the "unique up to isomorphism" be safely replaced with "unique up to bijection"?

Answer 1

The answer is no to all three questions.

1. $A$ and $B$ are isomorphic. You say that an object is unique up to isomorphism when it is completely determined by some (universal) property, up to isomorphisms (i.e. there could be another object with the same properties, but it would have to be isomorphic to the first one).
2. This question doesn't really make any sense to me.
3. This is a more interesting question. You are probably used to work in categories where the objects have underlying sets and where the morphisms are morphisms of the underlying sets. This is not always the case, so the answer is no. Also in the case where the objects have underlying sets and the morphisms are morphisms of sets respecting some additional structure, this is certainly not true (e.g. not all bijections between groups are isomorphisms...)