What's the disjoint union? I'm self-studying some analysis, and ran into the term 'disjoint union'. 
I googled it, and it seems that it's just a normal union of any sets, but where we pair each duplicate with an index indicating from which group it came from. So, basically, a disjoint union of $a = \{1,2\}$ and $b = \{2,3\}$ would become $\{1_a,2_a,2_b,3_b\}$.
Yet, the book says it's just a regular union, and the 'disjoint' part is just a way to let the reader know that the sets in question are disjoint (or, if the operator is used in a theorem, it implies that we require the sets to be disjoint).
So which one is it? It could be both, I figure, but would like to be sure.
 A: A disjoint union of $a = \{1,2\}$ and $b = \{2,3\}$ would be $\{1_a,2_a,2_b,3_b\}$.
In case the sets are already disjoint there is no need to mark the elements. So often this is the case that the author just wants to emphasise that the union is disjoint. In other cases you want to force the union to be disjoint. In those cases you would mark the elements like I did above.
A: I'll put this notion in a more general context:
The disjoint union of sets, is their coproduct in the category of sets, i.e. it is a solution (unique up to isomorphism) of the following universal problem:
Given two sets $A$ and $B$, find a set $C$ and two maps $i_A\colon A\to C$,  $i_B\colon B\to C$, such that for any set $X$ and maps $u_A\colon A\to X$, $u_B\colon B\to X$, there exists a map $f\colon C\to X$ such that
$$u_A=f\circ i_A,\quad u_B= f\circ i_B. $$
It is the basis for the topological sum of topological spaces, for glueing together topological spaces and for the construction of direct limits of sets or topological spaces.
