A question regarding interpreting the statement "the operation $A\amalg B$ is well-defined up to isomorphism"

Question

I have to prove that the operation $A\amalg B$ is well defined up to isomorphism.

I thought the phrase well-defined is associated only with mappings. OK let us assume the "operation" $A\amalg B$ is indeed the mapping of $A$ and $B$ to $A\amalg B$ by associating elements $a\in A$ with the tuple $(a,1)$ and associating $b\in B$ with $(b,1)$, with $A\amalg B$ containing all elements of the form $(a,0)$ and $(b,1)$.

How is this mapping well-defined up to isomorphism? Isn't the phrase "up to isomorphism" valid only for sets? Are we talking about the set $A\amalg B$ here or the mapping?

Answer 1

I suspect "well-defined up to isomorphism" is intended to mean "unique up to isomorphism" (as Zhen Lin has already commented).

For some very nice, helpful, remarks about how mathematicians are apt to rather mis-use the phrase "well-defined", see this blogpost by Tim Gowers.

Answer 2

Coproducts in categories are often denoted by $\coprod$. In general, the functor $(A,B) \mapsto A \coprod B$ is only defined up to (canonical) isomorphism. In a general category, there is no general canonical choice of the coproduct. Even in the category of sets, although there is a well-known model for $A \coprod B$, namely $A \times \{0\} \cup B \times \{1\}$, why is this canonical? Why not taking $A \times \{\pi\} \cup B \times \{\heartsuit\}$?