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 $|N{\mathbb{N}}|\cong |NJ[S^0]|\simeq B\Omega\Sigma S^\simeq |\Sigma S^0|\simeq S^1$.


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 $S^1$.

  • $\begingroup$ Your first method is interesting. Most other answers show that $N{\mathbb{N}}\to N{\mathbb{Z}}$ is a weak equivalence. However, you show that a weak equivalence can be found in the other direction. That is, you identify an embedding of the simplicial circle $S^1\hookrightarrow N{\mathbb{N}}$ that is a weak equivalence. Thank you! $\endgroup$ – Gao 2Man Apr 21 '12 at 4:34

Indirect method:

Monoids are categories with one object. A simple calculation shows that the inclusion of categories N into Z has contractible homotopy fiber (which is the nerve of the category whose objects are the arrows of the one object category Z and whose arrows are commutative triangles of Z mediated by the elements of N). Thus Quillen's theorem A yields a homotopy equivalence of the corresponding nerves arising from the inclusion of underlying categories.

Direct method:

Consult this paper by Ken Brown: "The Geometry of Rewriting Systems"

You need only the simplest version of his method. With it one can show that the nerve of N and the nerve of Z have cellular models which differ only by collapses of simplices and thus have the same (simple) homotopy type.


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.


I am not sure if this is the type of direct proof that you are looking for, but here it goes. I will start with a more general theorem: Let $M$ be a CANCELABLE monoid, and $K$ be the left adjoint to the forgetful functor, $U:GROUP\rightarrow MONOID$. Then $BM$ is homotopy equivalent to $BK(M)$. The way I like to see this is to think of both monoids and (therefore groups) as categories. By $B$, I mean the nerve of the category that turns the category into a simplicial set. Now I find these sorts of nerve theorems are much easier to see in the world of simplicial sets. To see this particular theorem, it suffices to try to build a minimal fibration (that is a fibrant replacement). In the case of a monoid, all of the inner horns are filled, and we must only find out how to fill the outer horns. But these will just be adding in the inverse that are not yet in the monoid. Further the minimal fibration condition will ensure that horn fillers are unique. Essentially, what you are doing is to perform a geometric version of the $K$ functor in the category of simplicial sets. As an interesting exercise, it would be good to take the minimal fibration associated to $B\mathbb{N}$ and see that you get the simplicial set, $B\mathbb{Z}$.

Now for the James construction that you mention (even though this is not part of your question, it is worth mentioning) their is a simplicial set version of this called the Milnor FK construction. What you do is you start with a reduced simplicial set, $X$ (a simplicial set with one vertex). We then define a simplicial group called $FK(X)$ the n-th group is the free group on the elements of $X_n$ modulo the image of the iterated degeneracey, $s_0^n(pnt)$, where $pnt$ is the vertex.

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    $\begingroup$ You are right, the Monoid has to be cancleable. see www.math.leidenuniv.nl/scripties/LenzMaster.pdf Page 32 $\endgroup$ – Spice the Bird Apr 14 '12 at 7:39
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    $\begingroup$ Seems it was not a cancelable conjecture. $\endgroup$ – Zack Wolske Apr 14 '12 at 18:02
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    $\begingroup$ I think that if M is a cancelative monoid that has a group of left fractions then you can just use Quillen's A on the inclusion. $\endgroup$ – Benjamin Steinberg Apr 14 '12 at 21:45
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    $\begingroup$ I don't understand several things about this answer: (1) You seem to be saying you can get BKM from BM by just adding fillers of outer horns, but of course, once you've added a few inverses of elements of M you need to deal with their compositions too. (2) Even if you didn't just mean fillers for outer horns in the original BM, you still seem to be saying that you get BKM from BM by just adding simplices; but having left and right cancelation does not guarantee that M->KM is injective! $\endgroup$ – Omar Antolín-Camarena Feb 20 '13 at 23:14
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    $\begingroup$ (3) You talk about a minimal fibration, but I don't know which map you mean: there is only the constant map BZ -> BN (simplicial maps like that are just monoid homomorphisms Z->N), and the inclusion BN->BZ is not a fibration. (Also, a fibrant replacement for a simplicial set X is a Kan complex Y with an acyclic cofibration X -> Y.) Did you mean instead that you wanted the thing you get by adding simplices to BM to be a minimal simplicial set? $\endgroup$ – Omar Antolín-Camarena Feb 20 '13 at 23:17

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