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It seems like when we assume "niceness" in homotopy theory we assume that $X$ has the homotopy type of a CW complex, and in fiber bundle theory we assume that $X$ is paracompact. How do these two interact? Is any space with the homotopy type of a CW complex paracompact? (In particular, is $I^I$ paracompact?)

(CW complexes are always paracompact and Hausdorff. According to Milnor (http://www.jstor.org/stable/1993204) a paracompact space that is "equi locally convex" will have the homotopy type of a CW complex. Also according to that paper, if $X$ has the homotopy type of a CW complex and $K$ is actually a finite complex then $X^K$ has the homotopy type of a CW complex.)

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    $\begingroup$ Paracompactness isn't a homotopy invariant. But CW complexes are paracompact, this is a common lemma in most intro algebraic topology texts. Your question is perhaps most appropriate for the math.stackexchange website. $\endgroup$
    – Ryan Budney
    Commented Nov 4, 2010 at 18:56
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    $\begingroup$ I think this question is fine. Using Brown's representability theorem, we get, for example, that the functor which assigns a space the set of isomorphism classes of prinicipal G-bundles is representable over the homotopy categeory of CW-complexes. But also, we know that the classifying space of G represents this same functor on paracompact Hausdorff spaces. So he is asking what is going on here... $\endgroup$
    – David Carchedi
    Commented Nov 4, 2010 at 19:33
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    $\begingroup$ In the homotopy category of CW-complexes, every object is actually a CW complex, right? Because then principal G-bundles are represented by maps into BG. If we enlarge the class of objects to spaces with the homotopy type of a CW-complex, then each space X has a homotopy equivalence from a CW complex K, pulling back the bundle gives a bundle over K, and then we classify this with a map $K \rightarrow BG$. Composing maps, we get a "classifying map" $X \rightarrow BG$. The problem, though, is that pulling back the bundle over $BG$ to a bundle over X doesn't necessarily give an isomorphic bundle, $\endgroup$
    – Cary
    Commented Nov 4, 2010 at 20:35
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    $\begingroup$ since the usual proof that two bundles are isomorphic uses the bundle homotopy theorem, and this theorem needs paracompactness. $\endgroup$
    – Cary
    Commented Nov 4, 2010 at 20:36
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    $\begingroup$ On the other hand, given a G-bundle on a non-paracompact space, it is the pullback of the universal G-bundle if and only if it trivialises over a numerable cover. This is one place where you have to use a genuinely different Grothendieck topology on Top other than the one involving arbitrary open covers. The two coincide for paracompact spaces, so the difference is subtle and waved over by most authors. $\endgroup$
    – David Roberts ♦
    Commented Nov 4, 2010 at 22:37

2 Answers 2

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$I^I$ is paracompact. It is a theorem of O'Meara that for $X$ a separable metric space, and $Y$ a metric space, then $Y^X$ - with the compact-open topology - is paracompact. $I$ is certainly a separable metric space, so the result holds.

As to your first question, I doubt that every space of the homotopy type of a CW-complex is paracompact (but this is intuition only). Something like $\mathbb{R}^{\aleph_2}$ or a similarly large-dimensional, non-metrizable topological vector space might do the trick, as it is contractible, hence the homotopy type of a CW-complex.

Edit: For any uncountable index set $J$, consider $\mathbb{R}^J$ in the product topology. It isn't normal, so isn't paracompact.

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For the fact that CW complexes are paracompact, I think that the proof in this book is better than I have seen elsewhere:


\bib{frpi:cst}{book}{
    author={Fritsch, Rudolf},
    author={Piccinini, Renzo~A.},
     title={Cellular structures in topology},
    series={Cambridge studies in advanced mathematics},
 publisher={Cambridge University Press},
      date={1990},
    volume={19},
}
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