So I know that every coproduct is not a product, so I am misunderstanding some part of the definition of (co)products. Saying that $U$ is a coproduct (the disjoint union of $X_1$ and $X_2$ below) of objects $X_1$ and $X_2$ implies that there are insertion maps from $X_1$ and $X_2$ respectively to $U$ ($i_1$ and $i_2$). And for any other coproduct, say $Y$, of $X_1$ and $X_2$ also with such insertion maps ($f_1$ and $f_2$), there is a unique isomorphism $f$ from $U$ to $Y$. This isomorphism logically must be constructed by taking each element in $U$ that was inserted from X1 and applying the insertion map $i_1$ from $X_1$ to $U$ to it $f_1$ and likewise for any element of $X_2$. In algebra, isomorphisms are bijective homomorphisms, implying the isomorphism is invertable. Or using the category-theoretic definition, for any isomorphism, there must be another arrow which, when composed on the left and right of the original isomorphism, yields the identity arrows for the domain and codomain respectively. Either way, there should be an arrow from $Y$ back to $U$ in the coproduct. The only way to logically construct this isomorphism would be to apply the inverses of the insertion maps for this coproduct ($f_1$ and $f_2$). But if these arrows all have inverses, then they can be "flipped around" to become a product and since (co)products are unique up to isomorphism, this same situation holds for the original coproduct $U$.
Is my logical error the assumption that the invertability of (co)product isomorphisms implies the invertability of the corresponding insertion maps? Or have a erred elsewhere? Thank you for any help you can provide in pointing out the source of my basic misunderstanding.