Let $ \mathcal{A}$ be a small category and $ \mathbf{B}\mathcal{A}$ its geometric realisation. It is evident that $ \mathbf{B}\mathcal{A}$ and $ \mathbf{B}\mathcal{A}^\circ$ are homotopy equivalent, and in fact homeomorphic. However, can we find functors that realise this equivalence? This post summarises some informal notes I have written on this following D. Quillen's paper `Higher Algebraic K-Theory: I", so grab the notes or read the summary below:`

Given any functor $ f:\mathcal{A}\to\mathcal{B}$, and an object $ B\in\mathcal{B}$, we can consider the category $ f^{-1}(B)$ consisting of objects $ A\in \mathcal{A}$ such that $ f(A) = B$. The morphisms of $ f^{-1}(B)$ are defined to be all the morphisms that map to $ 1_B$ under $ f$. Let us apply this to the following situation:

Given any small (or skeletally small) category $ \mathcal{A}$, we can construct another category $ S(\mathcal{A})$ as follows: the objects of $ S(\mathcal{A})$ are the arrows $ X\to Y$ of $ \mathcal{A}$, and a morphism $ (X\to Y)\to (X'\to Y')$ is a pair of morphisms $ X'\to X$ and $ Y\to Y'$ in $ \mathcal{A}$ making the obvious square commute. Now, we can consider the functor $ s:S(\mathcal{A})\to \mathcal{A}$ sending the object $ X\to Y\in S(\mathcal{A})$ to the object $ X\in\mathcal{A}$.

Consider now $ s^{-1}(X)$. It is easy to see that this category has an initial object (what is it?)—and hence applying the geometric realisation functor we get a contractible space $ \mathbf{B}s^{-1}(X)$. Now the functor $ s$ in fact makes $ S(\mathcal{A})$ into a precofibered category over $ \mathcal{A}$, and hence these two results imply together that $ \mathbf{B}S(\mathcal{A})$ and $ \mathbf{B}\mathcal{A}$ are homotopy equivalent, the homotopy equivalence being induced by $ \mathbf{B}s$.

Similarly, there is a functor $ t:S(\mathcal{A})\to \mathcal{A}^\circ$ where $ \mathcal{A}^\circ$ is the opposite category. The functor $ t$ sends a morphism to its target. Playing the same game, we can deduce that the functor $ t$ also induces a homotopy equivalence between $ \mathbf{B}S(\mathcal{A})$ and $ \mathbf{B}\mathcal{A}^\circ$. This shows that $ \mathbf{B}\mathcal{A}^\circ$ and $ \mathbf{B}\mathcal{A}$ are in fact homotopy equivalent; although this is evident from the definition of $ \mathbf{B}$, the geometric realisation, this constructs specific functors that realise the homotopy equivalence.

Of course, we skipped over many details here, which is why I have written some notes explaining this in much greater details with complete proofs, relying only on Quillen's Theorem A and elementary properties of the geometric realisation functor. They are only available in PDF format, since the explanations required quite a few commutative diagrams that would be painful to reproduce here on this blog.