Disjoint union on bundles One month ago I started to study fibre bundles form Jefrrey Lee's book (chapter 6). Now, traying to undesrtand them well, I'm dealing with the generic concept of bundle defined in Theory of fibre bundles (Husemoller, 1993), but in the category of topological spaces (topological manifolds in fact).
Let $\pi:B\rightarrow M$ be a bundle ($B$ and $M$ topological spaces and $\pi$ continuous). I can define the equivalence relation on $B$
$$
b\sim b' \Longleftrightarrow \pi(b)=\pi(b')
$$
and then consider $E=B/\sim$. As a set-mapping, I'm sure $f:E\rightarrow M$ defined by $f([b])=\pi(b)$ is bijective (and in fact if I'm not wrong, by the quotient topology answer 2-property $f$ is a homeomorphism, since $\mu\circ f = \pi which is continuous).
Question 1. Wiki says that the disjoint union of two or more non-empty topological spaces is disconnected. However, $B$ can be a connected space. Is it really true?
Question 2. Suppose for each $p\in M$ I have a non-empty topological space $B_p$. Would the disjoint union $\coprod_{p\in M} B_p$ a total space of some bundle? What would the topology be? Because I don't undestrand very well the coproduct topology.
Thanks
PD: I'm a physicist and I don't know much about topology nor category theory, so maybe some of these questions are trivial (or very difficult, I don't know.) So that I want to apologize in advance.
 A: To answer Question 1, it is quite possible to write a connected space as a disjoint union, for example the connected space $\mathbb{R}^2$ is a disjoint union of its vertical lines
$$\mathbb{R}^2 = \coprod_{x \in \mathbb{R}} \{x\} \times \mathbb{R}
$$
The confusion here is simply one of terminology. In this example, the "disjoint union topology" on $\mathbb{R}^2$ (which I prefer to call the "discrete union topology to avoid these types of confusions) is not the same as the naturally given "Euclidean topology" on $\mathbb{R}^2$, namely the topology generated by the open balls in $\mathbb{R}^2$. And so although it is true that the disjoint union topology on $\mathbb{R}^2$ is disconnected, that fact is irrelevent to our understanding of the Euclidean topology. In a fiber bundle $\pi : B \to M$, the topologies on $B$ and $M$ are given to you, and the fact that $B$ is disconnected in some disjoint union topology is irrelevant to our understanding of the topology on the fiber bundle.
To answer Question 2, the disjoint union topology on $B$ is indeed the total space of some bundle, namely the bundle where the given topology on the base space $M$ has been discarded and replaced with the discrete topology on $M$ (that's part of why I like to refer to the "discrete union topology"). But, once again, that fact is irrelevant to our understanding of the given topologies on $B$ and $M$.
Let me remark that I'm unsure what you mean by the "coproduct topology". The definition of a fiber bundle imposes conditions on how the topologies of $B$ and $M$ are related, and of additional key importance is how those are related to the topology of the (as yet unnamed) fiber $F$. In my comments you indicated your familiarity with the definition of smooth manifolds in terms of coordinate charts (and overlap maps). The definition of a fiber bundle has a similar logical structure in terms of coordinate charts (and overlap maps). Roughly speaking, a chart in $B$ is a homeomorphism 
$$\pi^{-1}(U) \mapsto U \times F
$$
for some covering of $M$ by open sets $U$, where the product topology is imposed on $U \times F$. I won't say anything about the overlap maps, but these are the concepts one needs to wrap one's head around in order to understand the formal definition of fiber bundles.
