Noncommutative sets

Let $\Fin$ denote the skeleton category of a category of finite sets. This means that the objects in $\Fin$ are the sets $[n]=\{0,1\hdots, n\}$ and morphisms are arbitrary functions. Let $F'$ denote a category with the same objects, but whose morphisms satisfy $f(0)=0$. Then there is a following diagram of categories

$$\xymatrix{ \Dl^{op} \ar@{>->}[d] \ar[r] & \Dl S'^{op} \ar[r]... ...@{>->}[d] \\ \Dl C =\Dl C^{op} \ar[r] & \Dl S \ar[r] & \Fin }$$

For a set $[n]$ we have
$$\begin{align*} \Aut_{\Dl S}([n])&= S_{n+1}, \\ \Aut_{\Dl S'}([n]) &= S_n. \end{align*}$$
The top row of this diagram will correspond to Hochschild homology, and the bottom row to cyclic homology, which we will define in next chapter.

If $A$ is an algebra, then $[n]\mapsto A^{\ox(n+1)}$ is a well defined functor $\Dl S\to \Mod$.

$$A^{\ox2} \rightrightarrows A,\; a\ox b\mapsto ab,\; a\ox b\mapsto ba.$$

The two maps $d_1, d_0\:[1]\to [0]$ become the same in $\Fin$. If $A$ is commutative, then $[n]\to A^{\ox(n+1)}$ factors through $\Fin$.

Thus $\Dl S$ can be viewed as a category of noncommutative sets. It has a following description

$$\Ob(\Dl S)=\{[n]\}$$

$$\Mor_{\Dl S}([n], [m]) = \text{ set maps preserving the order on fibers }f^{-1}(i) \text{ for }i\in [m].$$