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By the definition of disjoint union, for example, for $X=\{1,2\}$, $Y=\{3,4\}$, we have $X\sqcup Y=\{(1,1),(2,1),(3,2),(4,2)\}$, the index set is $\{1,2\}$ here. Thus in the rigorous sense we have $(X\sqcup Y)\cap X=\varnothing$?

Firstly I think that the disjoint union is just treated as an equivalent class, but in this case we cannot do the operation like $\cup$ and $\cap$. The result depends on the choice of the representatives.

Someone tells me that when we consider the disjoint union we fix the index set, some equations like $A\sqcup B=A\sqcup B$ do not make sense when we lose the information of index sets. But how about the well-known proposition: $$A\sqcup B=A\cup B\quad iff\quad A\cap B=\varnothing$$

Asaf Karagila
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Phanpu
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  • I think we must understand equality in your well-known proposition as isomorphism. – FShrike Oct 24 '22 at 14:22
  • It’s also very common to (again by isomorphism) identify $X$ with its inclusion in the disjoint union, so that it makes sense to talk about intersection – FShrike Oct 24 '22 at 14:23
  • Notation means what the author intends it to mean. It is not uncommon to write $A\sqcup B$ to literally mean $A\cup B$ but with the choice of having written it with a square cup instead as just a notational nod to allude to the fact that $A\cap B=\emptyset$ – JMoravitz Oct 24 '22 at 14:36

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