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For the symmetric group $\Sigma_n$, you can take \begin{align*} E\Sigma_n &= \{\text{injective functions } \{1,\dotsc,n\}\to\mathbb{R}^\infty \} \\\\ B\Sigma_n &= \{\text{subsets of size $n$ in } \ …

[UPDATE: There were some mistakes in the first version. Here is a more careful account.] I'll work everywhere with CGWH spaces, so I have a Cartesian closed category. Note that $BLG$ is always path …

The space $S^\infty$ is actually homeomorphic to $\mathbb{R}^\infty$. To see this, put \begin{align*} B_n &= \{x\in\mathbb{R}^\infty\::\: \|x\|\leq n, x_k = 0\text{ for } k \geq n\} \\ C_n &= \{x\ …

The space $\Omega BG$ is, by categorical nonsense, a classifying space for principal $G$-bundles over $\Sigma X$ with a chosen trivialisation at the basepoint. Any such bundle can be trivialised over …

[This version has been updated in response to comments from the OP] Recall that $B$ gives an endofunctor of the category of abelian topological groups. We can apply $B$ to the obvious map $\mathbb{Z …

Firstly, yes, your examples are all correct. However, in example~(B) we just have $O(3)=\{\pm I\}\times SO(3)$ as groups, so your fibration is just the product of $$BSO(3)\xrightarrow{1}BSO(3)\to 1$ …

Think about the case where $\pi_0(G)=\mathbb{Z}$, so $B(\pi_0(G))=S^1$, so we have a fibre bundle $BG_0\to BG\to S^1$. In this case $G$ is always a semidirect product formed using an automorphism $\a …

On the other hand, if we define $BG$ by the usual simplicial construction, then the functor $B$ does indeed preserve pullbacks. Indeed, if we start with a pullback square $(G,H,J,K)$ then in the corr …

I'll assume that by $BG$ you mean ``any space of the form $E/G$, where $E$ is a contractible space with free $G$-action''. (The alternative would be to define $BG$ as the geometric realisation of a s …