I need some help in understanding the meaning of open sets in Sum of topologies.
I studied in Bourbaki's General Topology that:
The sum of topologies is the final topology defined on the sum of family of Sets $(X_i)_{i \in I}$, denoted by $X$, with respect to the canonical mappings $j_i: X_i \rightarrow X$. On identifying each $X_i$ with a subset of $X$ by means of $j_i$, he claims that any subset $A$ of $X$ is open if and only if each of the sets $A \;\cap X_i$ is open in each topological space $X_i, \;i \in I.$ Moreover, each set $X_i$ is open as well as closed in topology on $X$.
Here's how I think we could prove the if and only if statement about open sets:
$X_i$ is identified as $X_i \times \{i\} \subseteq X.$ Now let $A \subseteq X$ be open in $X$. Then by the definition of final topology, each $j_i^{-1}(A)$ should be open in $X_i\;$ i.e. $\;X_i \times \{i\}, i \in I.$ Thus considering $X_i$ as a subspace of $X, \;j_i^{-1}(A)$ is open in $X_i$ if and only if $A \,\cap X_i$ is open in $X_i$.
Further, let $A = X_i$. Then, $\;X_i \,\cap X_j = \begin{cases}
\phi & j \neq i \\
X_i & j=i
\end{cases}\quad$ because, $\;X_i \,\cap X_j= \phi$ as subspaces of $X$.
Hence, $X_i$ is closed in all $X_j,\, j \in I$. So, it is closed in $X$. Similar arguments hold for proving $X_i$ is open in $X$.
Please verify whether the solution is correct?
Also, I find it hard to digest that why we are identifying each $X_i$ with a subset of $X$ by means of $j_i$? This step seems forceful to me, taken just to give a convenient way to define open sets in $X$. It does not seem to have anything to do with the definition of final topology.
Any help will be greatly appreciated.
Best Answer
Your solution as to why the $j_i[X_i]$ are open in the sum $X$ is correct: $j_i^{-1}[X_i]$ is either $X_i$ itself or empty, so open in $X_i$.
The whole point of the construction is to make the copies of the $X_i$ all disjoint. Consider e.g. a sum of countably many copies of $[0,1]$ (in the usual topology). We cannot take a union because then we get just $[0,1]$, we need separate copies of $[0,1]$ so the trick is to use the index set and have a map
$j_n: [0,1] \to [0,1]_n:= [0,1] \times \{n\}, j_n(x)=(x,n)$. The sets $[0,1]\times \{n\}$ are disjoint for different $n$, so now we can talk about copy $0$ of $[0,1]$, copy $1$ and so on, and we can unambiguously define $X = \bigcup_{n \in \Bbb N} [0,1]_n$, and give the sum/union the final topology w.r.t. the maps $j_n$ into $X$. Because we use the final topology, we can prove the universal property we need for a sum: having continuous maps $f_n: [0,1] \to Y$ to some space $Y$, for all $n$, we can define a unique map $f: X \to Y$ "by component", namely such that $f \circ j_n = f_n$ for all $n$. It's entirely dual to the product construction. In general, having made the $X_i$ disjoint (one can also show all $j_i$ are in fact homeomorphisms, so we have "copies" of each $X_i$ inside $X$) there never is any conflict between copies in defining the sum map.
The sum construction isn't very important in general topology, though it can be a handy tool for some proofs and examples, sometimes.