Confusion on Nerve of a Category and Segal's model of Classifying spaces

Question

If we have a topological category and the underlying category forgetting the topological structure, are the nerves same. They should be, is what my guess is from the definition of nerve of a category. Then, I have some facts which leads to contradiction (which should not be, thus I am missing something). The facts are below:

(For a category C, classifying space of C is the realization of the nerve of C.)

1. If G is a group consider it a category with the only object is * and set of morphisms is G. Then the nerve of this category has realization K(G,1).

2. If G is a topological group, then the classifying space (for Principal G-bundle) is not always (weakly-)equivalent to K(G,1). For example, take unit circle. They are same (upto weak-equivalence) when G has discrete topology.

3. In the paper of Graeme Segal (https://www.maths.ed.ac.uk/~v1ranick/papers/segalclass.pdf) his model of classifying space (in the sense of principal G-bundles) is in terms of realization of the nerve if the category in 1 (assuming say G is a locally finite CW-complex).

   Now, any two classifying spaces (in the sense of Principal G-bundle) are weakly equivalent, then 2 gives contradiction to 1 and 3 . I know I am mistaken at some stage. Any help is welcome.

Answer 1

I am sure that Segal does not mean them to be the same. In the cited paper, notice that Segal deliberately mentions "semi-simplicial spaces" (in modern terminology: "simplicial spaces") and their realisations. From that one infers that the nerve of a topological category is a simplicial space. By contrast, the nerve of an ordinary category is a merely a simplicial set, which one may regard as a simplicial space that is degreewise discrete.

Answer 2

No, these two nerves will not be the same. For example, let $G$ be a topological group. We can form its classifying category $BG$, and this is a topological category. If $G^\partial$ denotes the discrete group $G$, because we have a map $G^\partial \rightarrow G$ we get a map of classifying categories $BG^\partial \rightarrow BG$. This is a specific example of your comparison. This will almost never induce an equivalence on nerves.

Concretely, let $G=\mathbb{R}$. Then since the nerve of $BG$ is a model for the classifying space of $\mathbb{R}$, we see that it is contractible since all principal $\mathbb{R}$ bundles are trivial (alternatively you can see it has trivial homotopy groups).

However, $G^\partial = \mathbb{R}^\partial$ has $BG^\partial$ very far from trivial. It's fundamental group is uncountable!