Is there a direct way to seeing that $B{\mathbb{N}}\simeq S^1$, i.e. the classifying space of the monoid of natural numbers is homotopy equivalent to the circle? 


Here, since the natural numbers ${\mathbb{N}}$ is not a group, some care is needed to define the classifying space $B{\mathbb{N}}$ properly. One way to do this is to consider ${\mathbb{N}}$ as a discrete simplicial monoid, then set $B{\mathbb{N}}:=|N{\mathbb{N}}|$ to be the geometric realization of its nerve.


This fact is a special case of (yet surprisingly, equivalent to) a larger theorem of James, namely that James' construction $J[X]$ on a pointed simplicial set $X$ is weakly equivalent to $\Omega\Sigma |X|$. Here $J[X]$ is the free simplicial monoid on $X$ modulo the basepoint $\*$. Take $X=S^0$ gives $|N{\mathbb{N}}|\cong |NJ[S^0]|\simeq B\Omega\Sigma S^\simeq |\Sigma S^0|\simeq S^1$. 



Why is the classifying space of the natural numbers homotopy equivalent to the circle?
Method I: Symmetric products.


Contained inside the simplicial set $N\mathbb{N}$ is a copy of the simplicial circle $S^1$, generated by the zero-simplex and the 1-simplex $[1]$. This consists of all simplices of the form $e\_i = (0,\ldots,0,1,0,\ldots,0)$, together with the basepoint $(0,\cdots,0)$, in the simplicial object.


Moreover, $N\mathbb{N}$ is, levelwise, a commutative monoid, and the face and degeneracy maps are maps of commutative monoids. In fact, $N\mathbb{N}$ visibly is, in level $p$, the free commutative monoid on $e\_1, \ldots, e\_p$, or the infinite symmetric product of the based set $(S^1)\_p \subset (N\mathbb{N})\_p$. As a simplicial set, then, $N\mathbb{N}$ is the infinite symmetric product of the based simplicial set $S^1$.


Geometric realization preserves finite products and quotients by group actions (hence symmetric products), as well as colimits, so the geometric realization is homeomorphic to the map $S^1 \to Sym^\infty S^1$ of topological spaces. On homotopy groups, by the Dold-Thom theorem, this is the map $\pi\_\* S^1 \to H\_\* S^1$, which is known to be an isomorphism.


Method II: Covering spaces.


Consider the auxiliary simplicial set $E$, which is the nerve of the poset $\mathbb{Z}$ under $\leq$. $E$ is contractible, for example because the functions $f(x) \equiv 0$ and $g(x) = max(x,0)$ satisfy $f(x) \leq g(x) \geq id(x)$; these inequalities give rise to natural transformations of categories and thus a two-stage homotopy from the identity to a trivial map.


The group $\mathbb{Z}$ acts on $E$ freely (and properly discontinuously on geometric realization) by translation. I claim that the quotient is isomorphic to $N\mathbb{N}$. The p-simplices of $E$ are all of the form
$$
z \leq (z + n\_1) \leq \cdots \leq (z + n\_1 + \cdots + n\_p)
$$
and so the quotient can be identified with the collection of tuples $(n\_1,\ldots,n\_p)$. Composition adds adjacent $n\_i$ and inserting an identity inserts $0$, so this really is the simplicial set $N\mathbb{N}$.


Since geometric realization preserves quotients by group actions, this makes $B\mathbb{N}$ into a $K(\mathbb{Z},1)$, and hence homotopy equivalent to $S^1$.


