what is the classifying space of a monoid

Question

In the paper Homology Fibrations and the "Group-Completion". Theorem. McDuff, D.; Segal, G., 1976, the first line: A topological monoid $M$ has a classifying space $BM$.

I do not understand this sentence. Why a topological monoid $M$ has a classifying space $BM$? I find www.math.leidenuniv.nl/scripties/LenzMaster.pdf, but do not understand what the author talks about.

A classifying space $BG$ of a topological group $G$ is the quotient of a weakly contractible space $EG$ (i.e. a topological space for which all its homotopy groups are trivial) by a free action of $G$. It has the property that any principal $G$-bundle over a paracompact manifold is isomorphic to a pullback of the principal bundle $EG → BG$.

If we change "topological group" to "topological monoid", the definition of classifying spaces does not work.

Let $X$ be a pointed space, $\Sigma$ be the suspension, and the monoid $M$ be the labelled configuration space $C(\mathbb{R}^n;X)$. Then in order to obtain Corollary 3.3 of F. Cohen's paper: The homology of C n+1 spaces, n>=0, we need to prove $$BC(\mathbb{R}^n;X)\simeq \Omega^{n-1}\Sigma^nX.$$ How to obtain this equation?

Given $M$ as a monoid or a monoid up to homotopy, is there any procedure or method to follow how to get $BM$?

Answer 1

• Every monoid is a category with one object.
• Every small category has a classifying space, defined as the geometric realization of the nerve.
• The classifying space of a monoid is (by definition) the classifying space of the corresponding category.

Explicitly, if $M=(X,\cdot,1)$ is a monoid, its classifying space $BM$ is a CW-complex which has an $n$-cell for each element of $X^n$, and these are glued along the boundary maps $X^{n+1} \to X^n$, $(x_0,\dotsc,x_n) \mapsto (x_0,\dotsc,x_{i-1},x_i \cdot x_{i+1},x_{i+2},\dotsc,x_n)$.

In case of topological monoids, work with topological categories and their geometric realization.