When are principal bundles preserved by colimits?

Question

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.

5. Did I say anything wrong?

Answer 1

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.)