Why isn't every coproduct a product (and vice-versa)? 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.

 A: ShadesOfGrey you ask us to point out a logical error in your reasoning. Well - in a sense - there is no logical error. Your reasoning is perfectly correct! Look at it this way. 
1- You make the extravagant claim:
"The only way to logically construct this isomorphism would be to apply the inverses of the insertion maps for this coproduct (f1 and f2)."
then
2- you quickly and correctly arrive at a contradiction:
"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" 
So...bravo! You have made a perfectly acceptable proof by contradction that
"The only way to logically construct this isomorphism would be to apply the inverses of the insertion maps for this coproduct (f1 and f2)."
cannot be true! (did you notice the bold face?)
In other words: applying "the inverses of the insertion maps " is one possible way to "logically construct this isomorphism", but clearly cannot be the only way. Indeed in most cases this inversion is not possible (but in some categories/cases it is, and then products coincide with coproducts in these categories/cases). 
So, how do you "construct" the isomorphism $f$? Well, category theory does not tell you how, in general (as far as I know). It just tells you that there exists a unique iso $f$, but it does not tell you how to build it. It gives you a non-constructive proof. It just tells you that $f = [f_1,f_2]$ where the "[-,-]" is a special operation, but the constructing details depend on the specific category and are not given in general (as far as I know). 
You ask us if you erred elsewhere. In my opinion, if anything, you made an error in attitude. Please, let me explain.
You state boldly and with "Gaussian" certainty that 
"The only way to logically construct this isomorphism would be to apply the inverses of the insertion maps for this coproduct (f1 and f2)."
No hint of reasoning/proof is given for this extravagant claim. Do you realize that stating the existence of something in mathematics require some effort/proof? and that unique existence requires even more effort/proof? None of this is given (and cannot be given in this case).
Indeed your initial certainty crumbles a few lines later when you correctly find a contradiction. But instead of recognizing this as a legitimate proof by contradiction, you start questioning your reasoning and yourself.
Conclusion: if you make a reasoning and you arrive at a contradiction, do not question yourself. Question the statements you made in the course of the reasoning/proof. Including the assumptions. Check every word, adjective, comma, quantifier, ect. Write the statements in your own language to be sure. Even better write the statements in a formal language if possible (symbolic logic for example).   
In this case the "wrong words" were "the only" in the statement
"The only way to logically construct this isomorphism would be to apply the inverses of the insertion maps for this coproduct (f1 and f2)."
and should be replaced by "One possible". That's all.
A: The map $f$ is not necessarily an isomorphism.  If you're new to category theory, remember that "morphism" ≠ "isomorphism."
For example, in the category of sets, suppose that $X_1$ and $X_2$ are sets, each with two elements. Then $X_1 \coprod X_2$ is a set with four elements. Let $Y$ be a set with fewer than four elements. Then $f$ cannot be an isomorphism, because in the category of sets isomorphisms are bijections. 
Edit: In light of magma's excellent answer and helpful comment, I see that I was not answering the question asked.  Thanks Magma!
A: If $Y$ is a coproduct then it is certainly true, as you say, that there is a unique isomorphism $f$ that makes your diagram commute.
However nothing about this situation implies that $f_1$ or $f_2$ is an isomorphism. In other words, you are right to suspect this part of your reasoning.
It may be instructive to consider a concrete example. Suppose we’re in the category of sets, and $X_1$ and $X_2$ are both sets with one element, so $U$ and $Y$ are sets that have two elements. These sets are certainly isomorphic, but none of the insertion maps is invertible. (There can’t possibly be an isomorphism between a one-element and a two-element set, so there’s no way they could be invertible.)
