# Free groupoid and homotopy equivalence

Let $C$ be a (small) category. One can form the free groupoid $GC$ of $C$ which is the left adjoint construction to the inclusion functor $\mathrm{Groupoid}\rightarrow\mathrm{Category}$. Is then $C$ always homotopy equivalent to $GC$? In other words, are the spaces $BC$ and $B\pi BC$ homotopy equivalent where $B\underline{}$ is the classifying space and $\pi\underline{}$ is the fundamental groupoid?

• Which adjoint? The use of the word "free" suggests you mean the left adjoint, but in fact there is also a right adjoint (maximal subgroupoid). – Zhen Lin Jun 7 '13 at 16:13
• Yes the word "free" pins it down: It's the left adjoint. In other words, you localize the category at all morphisms. I also changed the word "forgetful" to "inclusion". – Werner Thumann Jun 7 '13 at 18:03

Basically take the idempotent semigroup with elements $(a,b)$ with a,b either 0 or 1 and multiplication is (a,b)(c,d)=(a,d). Next add an identity. Clearly the fundamental group of this idempotent monoid is trivial. It is known to have classifying space homotopic to a 2-sphere.
• Thank you for your answer. Do you know whether the Ore condition on $C$ is enough for the statement to be true (you mentioned in the other question that it's true in the case of one object categories). Furthermore, do you even know of an easy to verify condition on $C$ which is equivalent to $C\rightarrow GC$ being an homotopy equivalence? – Werner Thumann Jun 7 '13 at 15:06
• In fact, you can realize any homotopy type as a nerve of a category. Just take a closed under intersections cover $U$ of topological space $X$ by contractible subspaces. Then the nerve of corresponding poset is equivalent to $X$. On the other hand, groupoids can model only homotopy 1-types, i.e. spaces with $\pi_i (X)=0$ for $i>1$. To get unrestricted equivalence, one needs to consider higher categories and $\infty$-groupoids. – Anton Fetisov Jun 7 '13 at 21:43