Let $G$ be a topological group and consider a family $$G\rightarrow E_i\rightarrow B_i$$ of $G$-principal bundles indexed over the natural numbers. Suppose we have $G$-bundle morphisms (equivariant bundle maps) $$\require{AMScd}\begin{CD} G @>{=}>> G\\ @VVV @VVV \\ E_i @>{f_i}>> E_{i+1}\\ @VVV @VVV \\ B_i @>{g_i}>> B_{i+1}\\ \end{CD},$$

  1. Is the colimit $$\operatorname{colim}_{i} E_i\rightarrow \operatorname{colim}_{i} B_i$$ always a $G$-principal bundle if one does not assume any topological restrictions?
  2. Are there topological restrictions, such that 1) is true? Maybe some, which include all the examples I gave? A possible option would be to impose the maps $E_i\rightarrow E_{i+1}$ to be (closed) inclusion. This should be the case for all examples given.
  3. Does it hold it the category of compactly generated Hausdorff spaces?
  4. Does it hold in the category of compactly generated weak Hausdorff spaces?

This situation occurs quite often in algebraic topology, especially to find models for classifying spaces:

  1. $S^0\rightarrow S^{(n+1)-1}\rightarrow \mathbb{R}P^n$ yields $S^0\rightarrow S^\infty\rightarrow \mathbb{R}P^\infty$ and therefore $BS^0\simeq \mathbb{R}P^\infty$.
  2. $S^1\rightarrow S^{2(n+1)-1}\rightarrow \mathbb{C}P^n$ yields $S^1\rightarrow S^\infty\rightarrow \mathbb{R}P^\infty$ and therefore $BS^1\simeq \mathbb{C}P^\infty$.
  3. $S^3\rightarrow S^{4(n+1)-1}\rightarrow \mathbb{H}P^n$ yields $S^3\rightarrow S^\infty\rightarrow \mathbb{H}P^\infty$ and therefore $BS^3\simeq \mathbb{H}P^\infty$.
  4. $\operatorname{Diff}(M)\rightarrow \operatorname{Emb}(M,\mathbb{R^n})\rightarrow \operatorname{Emb}(M,\mathbb{R^n})/\operatorname{Diff}(M)$ yields $\operatorname{Diff}(M)\rightarrow \operatorname{Emb}(M,\mathbb{R^\infty})\rightarrow \operatorname{Emb}(M,\mathbb{R^\infty})/\operatorname{Diff}(M)$ and therefore $B\operatorname{Diff}(M)\simeq \operatorname{Emb}(M,\mathbb{R^\infty})/\operatorname{Diff}(M)$ for a compact smooth manifold.

I want to put these examples in a general setting. To conclude statements about the classifying space, one needs to know that the colimit is weakly contractible, which is mostly implied by the increasing connectivity of the spaces $E_i$. For sure, one has to impose topological restrictions to conclude, that the filtered colimit commutes with homotopy groups, e.g. that the spaces $E_i$ are compactly generated spaces and the maps $f_i\colon E_i\rightarrow E_{i+1}$ are closed inclusions.

  1. Did I say anything wrong?
  • $\begingroup$ Do you have a reference for the exactness of filtered colimits in the category of topological spaces? Does it hold in CGWH and CGH, too? $\endgroup$
    – Tom
    Apr 28 '15 at 14:29
  • $\begingroup$ I doubt this proves the claim. It just proves the underlying sets are isomorphic, not that the topologies coincide. $\endgroup$
    – Tom
    Apr 28 '15 at 14:38
  • $\begingroup$ Sorry it's actually not true that the forgetful functor creates limits, so the argument breaks there. $\endgroup$ Apr 28 '15 at 15:28

I came up with the following example which disproves what you hoped to be true: Take $G = \mathbb Z/2$. A $G$-principal bundle is the same as a two-fold covering. Take $p \colon S^1 \rightarrow S^1$ to be the two-fold covering map $z \mapsto z^2$. Now take $E_i \rightarrow B_i$ to be equal to $p$ and also all $f_i$ and $g_i$ to be equal to $p$.

Then one can see easily that the induced map $\text{colim}_{i \rightarrow \infty} E_i \rightarrow \text{colim}_{i \rightarrow \infty} B_i$ is not a two-fold covering map: Indeed, this map is $\mathbb R/\mathbb Z[1/2] \rightarrow \mathbb R/\mathbb Z[1/2], [x] \mapsto [2x]$ so the basepoint $[0]$ does not have two preimages.

One might wonder what happens when one replaces the (in general quite badly behaved) colimit by the homotopy colimit. But it turns out that the induced map $\text{hocolim}_{i \rightarrow \infty} E_i \rightarrow \text{hocolim}_{i \rightarrow \infty} B_i$ is still not a two-fold covering: Take the same example as above, then this map is $B(\cdot 2)\colon B(\mathbb Z[1/2]) \rightarrow B(\mathbb Z[1/2])$ which induces an isomorphism on the fundamental group, hence can not be a two-fold covering.

(One can also take the mapping telescope to model the homotopy colimit and see quite explicitly that the map is not a covering map.)

  • $\begingroup$ Unfortunately, this does not answer my question, since $p\colon S^1\rightarrow S^1,z\mapsto z^2$ is not equivariant and hence $p$ not a $G$-bundle morphism. $\endgroup$
    – Tom
    Apr 28 '15 at 8:18
  • 5
    $\begingroup$ Modify the example so that $f_i$ and $g_i$ are given by $z\mapsto z^3$. The colimit of $B_i$ is a space in which every point is dense, so it cannot have a nontrivial covering space. $\endgroup$ Apr 28 '15 at 14:54
  • $\begingroup$ What if I impose the maps $E_i\rightarrow E_{i+1}$ to be (closed) inclusions? $\endgroup$
    – Tom
    May 1 '15 at 9:40

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