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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.

Dog_69
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  • Regarding Question 2, I can't stand the terminology of the "disjoint union of topological spaces", because it causes exactly the confusion you've perceived. "Discrete union" is a much better terminology. But the point is that even though $B$ is the set theoretic disjoint union of its fibers, it is not the topological disjoint union, i.e. it is not the discrete union. – Lee Mosher Dec 16 '17 at 04:50
  • Think, for example, $B=\mathbb{R}^2 \mapsto E=\mathbb{R}$ by projecting onto the $x$-axis. The fibers are the vertical lines in $\mathbb{R}^2$, which is the set theoretic disjoint union of its vertical lines but is not the topological disjoint union. – Lee Mosher Dec 16 '17 at 04:51
  • So the problem is the topology, I suppose. And can some homeomorphism $h:B\rightarrow \coprod_{p\in M}\pi^{-1}(p)$ exist? I mean, Can I relate two topologies in some way? – Dog_69 Dec 16 '17 at 11:21
  • The sets $B$ and $\coprod_{p \in M} \pi^{-1}(p)$ are the same set. So there are not two topologies, there is only one topology on that one set. Also, usually one doesn't bother with $B / \sim$ since, as you point out, it corresponds bijectively with $M$ itself. I can try to write an answer, but it would greatly help to know how much topology you know? Do you know the definition of a topology? The definition of a basis of a topology? Do you know the definition of a smooth manifold, expressed in terms of coordinate charts and smoothness of overlap maps? – Lee Mosher Dec 16 '17 at 14:05
  • I know these three concepts. – Dog_69 Dec 16 '17 at 20:27

1 Answers1

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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.

Lee Mosher
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  • I hace studied fibre bundles un terms of local charts, since is the approach followed by Jeffrey Lee. But Husemoller doesn't require this locality. For his, a bundle is a general concept which involves 3 elements: $B,M$ and $\pi$ and that can occur not only in topological spaces (smooth manifolds) but also in Category theory. And that is the reason I wanted to try to understand them. Because bundles and fiebre bundles can be understood in a more general way, and I think it is convinient for me to know it. – Dog_69 Dec 17 '17 at 17:31
  • Although Husemoller does not require local triviality throughout his book, in my opinion that is done primarily for pedagogical reasons. You'll see several key places in his book where local triviality becomes important. You could probably read the whole book with local triviality assumed and not miss anything particularly important. – Lee Mosher Dec 17 '17 at 21:30
  • The last question @Lee. Let $M$ be a manifold and suppose that for each $p\in M$ there is attached a vector space $E_p$, all them homemorphic to a fixed vector space $V$. Let $E=\coprod_{p\in M}E_p$. Is there a natural way to define a vector bundle structure on $E$? Or I need also to define a cocycle ${\rho_{i,j}}$ on $M$? – Dog_69 Dec 22 '17 at 11:08
  • There has to be some additional data. It could of course be the defining data where $E$ is given a topology and an atlas of locally trivial charts as in the definition. Otherwise one could do something entirely arbitrary and un-natural: "for each $E_p$ choose a vector space isomorphism with ${p} \times \mathbb{R}^n$, and therefore $E$ is isomorphic to the trivial bundle $M \times \mathbb{R}^n$"; or "for each $E_p$ choose a vector space isomorphism with $T_p M$, and therefore $E$ is isomorphic to the tangent bundle $TM$"; but in general $E$ cannot be isomorphic to both! – Lee Mosher Dec 22 '17 at 16:03