I'm having trouble understanding what classifying spaces are in general.
It seems to me, that they are terminal in a category of bundles whose morphisms are pullback squares, is this correct?
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$\begingroup$ You have been asking many questions recently. Perhaps it would be better if you slowed down a little, and invested the time and effort necessary to make the questions good. We have a "how to ask" page (linked at the top) with some information about writing questions that are clear and motivated. $\endgroup$– S. Carnahan ♦Commented Dec 3, 2012 at 17:19
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2 Answers
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Actually a classifying space $BG$ for the group $G$ is weakly terminal (existence but not uniqueness) in the category of principal $G$-bundles with commuting squares for maps. Sometimes this sort of thing is called 'generic' in category theory, in this case the generic $G$-torsor. Also, depending on which model you take, it may not classify all bundles, for example Segal's model for the classifying space of a well-pointed topological group only classifies numerable bundles: those trivialisable over an open cover with a subordinate partition of unity. Off the top of my head I can't think of a universal bundle that classifies all principal bundles (such a thing would have to not be numerable itself, as Segal's and Milnor's constructions are).
Note that given a collection of transition maps $U_{ij} \to G$ over a non-numerable cover $U = \lbrace U_i \rbrace$ of some space $X$ ($G$ well-pointed, here), there is another reason you cannot get a classifying map $X\to BG$, namely that the canonical map $|NU| \to X$, where $|NU|$ is the geometric realisation of the nerve of the topological groupoid associated to $U$, is not split, as it is in the case when $U$ is numerable, though it is a weak homotopy equivalence. This raises the question of whether the functor $X \mapsto GBund_X$ is representable on the homotopy category of $Top$, rather than on the category of spaces and homotopy classes of maps. This depends on whether or not an arbitrary space is weakly equivalent to a paracompact space.
answered Dec 2, 2012 at 21:34
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$\begingroup$ but definitely commuting squares as morphisms, not pullback squares? $\endgroup$ Commented Dec 2, 2012 at 21:38
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$\begingroup$ Commuting squares with vertical arrows principal $G$-bundles are pullback squares; this relies on the theorem that any map between principal bundles over the same base is automatically an isomorphism. $\endgroup$– David Roberts ♦Commented Dec 2, 2012 at 21:51
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1$\begingroup$ CW complexes are paracompact, so any space is weakly equivalent to a paracompact space. See Fritsch and Piccinini ``Cellular structures in topology'', page 29 for a proof. It is not unreasonable to redefine principal bundles by restricting attention to numerable bundles (even without assuming paracompact base spaces), especially since the known universal bundles are numerable, hence so is any bundle pulled back from such a universal bundle. $\endgroup$ Commented Dec 2, 2012 at 23:35
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$\begingroup$ Ah, thanks, Peter. I knew there was a simple resolution to my final sentence. I was going to mention that the known universal bundles are numerable, hence everything classified by them is, but it seemed labouring the point, as I implicitly said it. I agree that restricting attention to numerable covers is the best resolution to this apparent dilemma, as non-numerable covers don't seem to occur in practice. $\endgroup$– David Roberts ♦Commented Dec 2, 2012 at 23:44
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Yes. With morphisms up to homotopy.
answered Dec 2, 2012 at 21:17
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$\begingroup$ I don't understand the second sentence. I'm thinking of bundles as simply morphisms in some category. $\endgroup$ Commented Dec 2, 2012 at 21:21