Complex topological K-theory - the book

Question

I am reading the first chapter of the book "Complex topological K-theory" by Efton Park, which is in general very good. However, for some reasons, which I don't understand, when working with compact Hausdorff spaces, he assumes that the space has finitely many connected components. This is an implicit assumption e.g. in the proof of Proposition 1.5.19, or explicit in the proof of Proposition 1.7.9.

Is there some reason for this restriction, or is the book seriously flawed?

EDIT: Let me state as an example the first of these propositions: Let $A$ be a closed subspace of a compact Hausdorff space $X$, let $V$ and $W$ be vector bundles over $X$, and suppose $\sigma: V\upharpoonright A\rightarrow W\upharpoonright A$ is a bundle homomorphism. Then $\sigma$ can be extended to a bundle homomorphism $\tilde{\sigma}: V\rightarrow W$.

The proof starts with the sentence: Without loss of generality, we assume that $X$ is connected; otherwise, work with $X$ one component at a time.

This is fine if $X$ has finitely many components. Not so fine if $X$ is e.g. the Cantor set and each component is a singleton.

EDIT2: The propositions mentioned above are (as indicated by the "e.g.") just examples. They are not the only affected results in the book. The same "problem" repeats throughout Sections 1.5 and 1.7 and I cannot state all the propositions here. So the question is really targeted to those who have and have read the book.

Answer 1

A bit late to answer but considering how there's no answer here I feel I should - the terminology of connected components is for sure technically incorrect (the example with the Cantor set is a good visual to show why), but the actual proofs are still fine because the only time we use "connected components" in proofs is to ensure constant dimension. So what are these connected components we're talking about if they're not what are traditionally called connected components?

Well, if $X$ is our base space and $\pi : V \rightarrow X$ is our bundle, by definition we have for each $x \in X$ some neighborhood $U_{x}$ with $V \vert U_{x} \cong \Theta^{n}(U_{x})$, so in particular it's easy to see the points $x$ such that $\text{rank}(V \vert U_{x}) = n$ is open - let's call this set $\text{rank}_{n}(X)$, to represent the collection of points of $X$ who have locally trivial neighborhoods of rank $n$. By definition of vector bundles, we know that $$ X = \bigcup\limits_{n=0}^{\infty}\text{rank}_{n}(X) $$ since every point has a locally trivial neighborhood. However, noting that each $\text{rank}_{n}(X)$ is open and disjoint from $\text{rank}_{m}(X)$ for $n \not= m$ (which is again easy to see), we have $$ X \setminus \text{rank}_{n}(X) = \bigcup\limits_{m\not= n}\text{rank}_{m}(X) $$ which is open, and thus each $\text{rank}_{n}(X)$ is clopen - in fact it is these sets which are being referred to as components! From here, the reason we can choose finitely many of them in the proofs comes down to the fact we're working over compact Hausdorff spaces (this is where we use the open in clopen), and when we work "inside" these "components" because they are closed subsets of a compact space they are also compact, and thus the usual "choose an open cover $\lbrace U_{1}, U_{2}, ..., U_{n} \rbrace$ such that $V \vert U_{i} \cong \Theta^{n}(U_{i})$..." argument holds (this is where we use the closed in clopen). Hope it clears up some understanding!