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Background. This question belongs to evil mathematics. It is motivated by this question which links to a paper in which it is claimed that it is an open problem whether there exists strictly functorial choices for pullbacks in the category of sets. While thinking about this problem, I had the idea to warm up with coproducts:

Definition. Let $C$ be a category with coproducts. Then for all objects $X,Y,Z$ we may choose the iterated coproducts $X \coprod (Y \coprod Z)$ and $(X \coprod Y) \coprod Z$. If we have choosen them, there is a canonical isomorphism $\alpha$ between these coproducts. Let's call $X,Y,Z$ strictly associative (with respect to coproducts) if the coproducts may be chosen in such a way that this isomorphism equals the identity. This means, in particular, that the objects $X \coprod (Y \coprod Z) = (X \coprod Y) \coprod Z$ are equal.

Question. In the category (Set) of sets and maps, is it true that every triple of objects is strictly associative? In other words (without category theoretic terms), is it possible to choose for every pair of sets $X,Y$ a set $X \coprod Y$ which is the union of two disjoint sets equipped with bijections to $X$ and $Y$, such that $X \coprod (Y \coprod Z) = (X \coprod Y) \coprod Z$ as a disjoint union of three sets equipped with bijections to $X$, $Y$ and $Z$?

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    The answer depends on your choice of coproduct, but I think it's likely to be no. The usual implementation of disjoint union using tagged unions (i.e. $A \amalg B = \{ 0 \} \times A \cup \{ 1 \} \times B$) is non-associative, for example.2011-12-06
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    Please read my question more carefully. Yes, it depends on the choice, and my question is exactly if particular "evil" choices are possible. The usual choice, of course, does not work.2011-12-06
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    I'm not sure about the answer at the moment, however the question is not clear to me. Do you want the following property: "For every two sets, $X$, $Y$ we have a disjoint ...", and you want this property when applied to triplets would be exactly the same? (That is the canonical choice of disjoint sets be uniform so it can be used in union of *three* sets)2011-12-06
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    @Zhen: This modified version also does not satisfy $(A \coprod B) \coprod C = A \coprod (B \coprod C)$ (and is no disjoint union, for example when $A=B$, as you've already mentioned). Perhaps we can make a case distinction ... Asaf: I have to admit that the details are a bit subtle. But I don't understand your suggestion. Could you please indicate what part of the question above is unclear?2011-12-06
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    https://arxiv.org/abs/1412.6714 Claims that we can choose strictly functorial pullbacks in Set.2016-11-06

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I'm going to cheat and assume the axiom of global choice. We well-order the universe of sets, so that every set $A$ is equipped with a well-ordering (of type $\alpha_A$). We then set $A \amalg B$ to be the von Neumann ordinal of $\alpha_A + \alpha_B$, and set $A \hookrightarrow A \amalg B$ and $B \hookrightarrow A \amalg B$ to be the obvious order-preserving maps. It is then clear that $A \amalg (B \amalg C) = (A \amalg B) \amalg C$ as sets (because ordinal addition is strictly associative) and they are even isomorphic as coproducts (because the insertions are the same).

Obviously, this fails to be commutative...

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    Ha, what a nice, nasty trick! Thank you. I'm OK with global AC. So the picture might be that we "flatten the universe" and then the coproduct is given by concatenation. // By "fails to be commutative" you obviously mean that it fails to be strictly commutative. Indeed, and this motivates another question: Is it possible to find coproducts in (Sets) which are strictly associative *and* strictly commutative?2011-12-06