TITLE: Why is the classifying space of the natural numbers homotopy equivalent to the circle?
QUESTION [21 upvotes]: 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 $|N{\mathbb{N}}|\cong |NJ[S^0]|\simeq B\Omega\Sigma S^\simeq |\Sigma S^0|\simeq S^1$.

REPLY [10 votes]: Applying the functor "free $\mathbb Z$-module", the abelian monoid becomes a commutative ring $\mathbb Z[T]$, a polynomial ring in one variable. The bar construction, a simplicial set, becomes a simplicial abelian group: the bar construction for the augmented $\mathbb Z$-algebra. It follows that the homology of the classifying space is $Tor^{\mathbb Z[T]}(\mathbb Z,\mathbb Z)$, which is the homology of the circle. That, plus the fact the fundamental group is what it should be, plus the fact that the classifying space inherits its own commutative monoid structure, gives the result.