Identification topology and disjoint unions I was reading the book Basic Topology by M.A. Armstrong and I came across something I couldn't understand. I have uploaded the relevant pages- 
(1)  What  exactly is a disjoint union and what is the topology on it? 
(2) Why is the function j continuous ?
(3)  The example related to Figure 4.2, is unclear to me. How does the subspace topology give a space homeomorphic to the circle , whereas the identification topology does not ? 
 A: I had the same questions while reading that particular page in Amstrong's book. It's a slight lack of clarity in an otherwise excellent book. Here is what I figured out:
A good introduction/explanation to/of disjoint union(s) is given in the Appendix (p. 145) of Lee's Introduction to Toplogical Manifolds. If you cannot get a copy of the book, I'd be happy to type in the relevant sentences from Lee's book. The important thing to understand is that the collection of sets in the disjoint union need not necessarily be disjoint, but still a disjoint union could be defined by "tagging" the index to each set. For example, it is possible to form a disjoint union consisting of "five copies of R", in which we may consider different copies to be disjoint from each other by tagging each element in each copy with elements from the set (1,2,3,4,5}.
A topology is defined on a disjoint union as follows: $O$ is an open set in the disjoint union $\sqcup X_\alpha$ if and only if $O$'s intersection with each $X_\alpha$ is open. This topology is called the disjoint union topology. From this definition it is clear that the inclusion map  $j: X_\alpha \rightarrow \sqcup_{\alpha}{X_\alpha}$ is continuous, since $j^{-1}(O) = O \cap X_{\alpha}$. 
Hope this helps.
