# Why isn't the ∞-groupoidification of a category 1-truncated?

I don't know how to reconcile the following facts:

• The nerve $$N(\mathcal C)$$ of a category is a 2-coskeletal simplicial set (source).
• An $$n$$-coskeletal Kan complex has trivial homotopy groups above degree $$n-1$$ (source).
• The geometric realisation of a category is a Kan fibrant replacement of its nerve (identifying Kan complexes with topological spaces for my purposes) (source).
• The Kan fibrant replacement of the nerve of a category has the same geometric realisation as the nerve, up to weak homotopy equivalence (source). In particular, it should have the same homotopy groups.
• Yet, every space arises as the geometric realisation of a category (even a poset). In particular, this MSE answer gives the poset displayed below as an example of a category whose geometric realisation is a sphere (and I have no trouble believing this).

So, I apologise for asking such a vague question, but intuition is completely failing me and I am very confused: how can the geometric realisation of a 1-category have higher homotopical information? Where does it come from? Is this related to model category theory, i.e. using 1-categories to model ∞-categories?

The mistake lies in combining the first two bullets to conclude that $$N(\mathcal{C})$$ has no nontrivial higher homotopy groups, because $$N(\mathcal{C})$$ is generally not a Kan complex*. On the other hand, a Kan fibrant replacement of $$N(\mathcal{C})$$ will generally not be $$2$$-coskeletal anymore. The rest of the bullets are all correct (the first two as well, for that matter), but since we have found we don't actually have any restrictions anymore on the higher homotopy groups of $$N(\mathcal{C})$$, you don't get a contradiction anymore.
*It is a Kan complex precisely if $$\mathcal{C}$$ is a groupoid. But in that case, $$N(\mathcal{C})$$ is a disjoint union of the nerves of connected components of $$\mathcal{C}$$, so let us assume for simplicity that $$\mathcal{C}$$ is connected. Then it is up to equivalence of categories a group, and hence $$N(\mathcal{C})$$ is a $$K(G,1)$$ for some group $$G$$. This indeed has no higher homotopy groups, so in those cases where our initial category $$\mathcal{C}$$ satisfies that $$N(\mathcal{C})$$ happens to be a Kan complex, it does hold that its geometric realization has no higher homotopy groups and we have no contradiction.