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This is a continuation of a previous question.

Is it possible to find a map $S : \mathrm{Set} \times \mathrm{Set} \to \mathrm{Set}$ such that $S(X,Y)$ is a coproduct of $X$ and $Y$ (thus it is equipped with universal morphisms $X \rightarrow S(X,Y) \leftarrow Y$; also $S$ will be a functor) and that $S$ is strictly commutative in the sense that the canonical bijection $S(X,Y) \cong S(Y,X)$ is the identity?

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Again, we abuse the axiom of global choice: we well-order the universe of sets, and set $A \amalg B$ to be the von Neumann ordinal $$\min \{ \alpha_A, \alpha_B \} + \max \{ \alpha_A, \alpha_B \}$$ where $\alpha_A$ is the order type of $A$. The coproduct insertions are the obvious order-preserving inclusions. A variation on this is to linearly order the universe; the point is that we want every pair of sets to be linearly ordered.

Note, however, that strict associativity fails now...

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    The keen observer will notice that almost always the result is just $\max\{\alpha_A,\alpha_B\}$...2011-12-08
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    @Asaf: Are you sure? This is ordinal arithmetic. at Zhen: Thanks. Could you explain what the coproduct injections are?2011-12-08
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    @Martin: $\omega+\omega_1=\omega_1$.2011-12-08
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    @Martin: If $\alpha_A < \alpha_B$, then inject $A$ into the initial segment $\alpha_A$ of the coproduct and inject $B$ into the remainder above that; vice-versa if $\alpha_A > \alpha_B$; and if $\alpha_A = \alpha_B$, tiebreak by considering which of $A$ or $B$ is greater in the ordering of the universe.2011-12-08
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    The last case causes problems for $A=B$, right? Also I don't see strict commutativity.2011-12-09
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    @Martin: If $A = B$ then it doesn't matter anyway. We have strict commutativity because the underlying set of $A \coprod B$ and $B \coprod A$ will always be equal and the insertions are canonical w.r.t. the well-ordering of the universe.2011-12-09