Suppose $G$ is a topological group and $H\leq G$ is a closed subgroup. The inclusion $H\to G$ induces a map on classifying spaces $BH\to BG$. I've seen in some sources that $BH\to BG$ is actually a fiber bundle with fiber $G/H$, for example in the following notes PDF, Proposition 2.15 on the bottom of page 59.
I understand the reasoning to be as follows. It's a general fact that $EG\to (EG)/H$ is a principal $H$-bundle, and since $EG$ is contractible, we know that $EG\to (EG)/H$ is the universal $G$-bundle. Therefore $(EG)/H$ is homotopy equivalent to $BH$.
It's also a general fact that $(EG)/H \to B$ is a fiber bundle with fiber $G/H$. However, I feel like we can't just replace $(EG)/H$ with $BH$ in this context, since they are only homotopy equivalent. And, in the PDF above, the author write $(EG)/H = BH$, which is perplexing.
Here's another confusion I have with this statement. Take $H$ to be the trivial group. Then $BH\to BG$ should be a fiber bundle for all (reasonable) topological groups $G$. But this fishy to me. I find it hard to believe that, at the very least, $BH$ surjects onto $BG$ for any $G$.
How can I reconcile these issues?
The statement "$BH \to BG$ is a fiber bundle" is a type error; "fiber bundle" is a word you apply to a map between topological spaces, but $BH$ and $BG$ are homotopy types. Being a fiber bundle is not a homotopy-invariant notion, and in fact every map between, say, connected CW complexes is a fiber bundle up to homotopy equivalence.
The homotopy-invariant statement is that the homotopy fiber of the map $BH \to BG$ is $G/H$; this statement does not depend on a choice of a point-set model for $BH$ or $BG$. An equivalent statement is that you can always pick a choice of model for $BH$ and $BG$ such that $BH \to BG$ (which here now refer to spaces) is a fiber bundle with fiber $G/H$.
When $H$ is the trivial group, $BH$ is (homotopy equivalent to) a point, and so we conclude that the map $\bullet \to BG$ has homotopy fiber $G$. It's more generally true that if $X$ is path-connected then the homotopy fiber of $\bullet \to X$ is the loop space $\Omega X$.