Contained inside the simplicial set $B\mathbb{N}$ 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,$B\mathbb{N}$N\mathbb{N}$ is, levelwise, a commutative monoid, and the face and degeneracy maps are maps of commutative monoids. In fact, $B\mathbb{N}$ 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 $(S^1)_p \subset (B\mathbb{N})_p$N\mathbb{N})_p$. As a simplicial set, then, $B\mathbb{N}$ 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$g(x) = min(x,0)$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 $B\mathbb{N}$. 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$B\mathbb{N}$.N\mathbb{N}$.
Since geometric realization preserves quotients by group actions, this makes $|B\mathbb{N}|$ B\mathbb{N}$into a$K(\mathbb{Z},1)$, and hence homotopy equivalent to$S^1$. 1 Method I: Symmetric products. Contained inside the simplicial set$B\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,$B\mathbb{N}$is, levelwise, a commutative monoid, and the face and degeneracy maps are maps of commutative monoids. In fact,$B\mathbb{N}$visibly is, in level$p$, the free commutative monoid on $e_1, \ldots, e_p$, or the infinite symmetric product of the based set $(S^1)_p \subset (B\mathbb{N})_p$. As a simplicial set, then,$B\mathbb{N}$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$g(x) = min(x,0)$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$B\mathbb{N}$. 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$B\mathbb{N}$. 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\$.