# A is Homotopy Equivalent to A^op via Functors

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.