# Classifying space of a monoid with 5 elements

In this paper https://arxiv.org/pdf/math/0202260.pdf, the author proves that $$BP$$ is homotopic to $$S^2$$, using homology (In the first lemma of the paper).

Well, I tried to use the definition of geometric realization to see how $$BP$$ looks like. I can see that everything just collapses into five 1 simplices via the multiplication defined. But I can't seem to connect these together, to form $$S^2$$.

What am I missing here?

• What do you mean by "everything collapses into five 1-simplices"? I don't know anything about monoids, but for groups, the classifying space can be obtained by looking at the geometric realization of the nerve of its one-element groupoid. In case this construction still works for monoids, then $BP$ should certainly have a $5$-fold wedge of circles as its $1$-skeleton, but the algebraic relations should also give rise to $2$-simplices glued appropriately. – Sofie Verbeek Oct 27 '18 at 13:09
• Because $x_{ij} x_{kl} = x_{il}$, any $n$ chain in $B_n(P)$ becomes a 1-chain. ----(1) So, it gives rise to five 1 simplices. And as you mentioned, it becomes a 5 fold wedge of circles. "...also give rise to 2-simplices glued appropriately" - I can't see how it should give rise to 2 simplices, because of (1). – wanderer Oct 27 '18 at 14:59