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Consider the classifying space of a category as defined in this post. Consider the case of a group $G$. I can't see something pretty simple, that is that this coincides with the 'standard' classifying space $BG$.

It seems to me that the Nerve of $C$ is the same as the contractible simplicial space $EG$ which covers $BG$: namely, $EG$ is constructed as having as $n-1$-simplices the $n$-uples of $G^n$, and then one quotients by the diagonal action of $G$, obtaining $BG$.

Now it would seem that the classifying space of the category $G$ (one only object, and one morphism for every element of $G$, interpreted as left multiplication) gives $EG$ and not $BG$. Why am I wrong? I suspect that the point comes when constructing the geometric realisation, but when?

Thank you in advance.

W.Rether
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1 Answers1

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Let $E$ be the category with object set $G$. Define $Hom_E(a,b) = \{(k,a,b): ak = b\}$. In other words, there is exactly one morphism between every object, represented by the group element that multiplies the domain to the codomain. Observe that $G$ acts on $E$ via $g \cdot (k,a,b) = (k,ga,gb)$.

  1. The quotient category $E/G$ is exactly the one-object category $\diamond$ associated to the group $G$.
  2. The $G$-action on $E$ is free. Therefore the induced action of $G$ on the classifying space $BE$ is also free. On the other hand, $BE$ is contractible because $E$ has an initial object (it is actually equivalent to a one object category with only the identity map). Thus the quotient $BE/G$ is a model for $K(G,1)$.

Thus $BE/G = B(E/G) = B (\diamond)$ is a $K(G,1)$ (the first equality can be verified looking at individual simplices).

Cihan
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