Example 60.19.1. Suppose that $A_*$ is any cosimplicial ring. Consider the cosimplicial module $M_*$ defined by the rule

For a map $f : [n] \to [m]$ define $M_*(f) : M_ n \to M_ m$ to be the unique $A_*(f)$-linear map which maps $e_ i$ to $e_{f(i)}$. We claim the identity on $M_*$ is homotopic to $0$. Namely, a homotopy is given by a map of cosimplicial modules

see Section 60.16. For $j \in \{ 0, \ldots , n + 1\} $ we let $\alpha ^ n_ j : [n] \to [1]$ be the map defined by $\alpha ^ n_ j(i) = 0 \Leftrightarrow i < j$. Then $\Delta [1]_ n = \{ \alpha ^ n_0, \ldots , \alpha ^ n_{n + 1}\} $ and correspondingly $\mathop{\mathrm{Hom}}\nolimits (\Delta [1], M_*)_ n = \prod _{j = 0, \ldots , n + 1} M_ n$, see Simplicial, Sections 14.26 and 14.28. Instead of using this product representation, we think of an element in $\mathop{\mathrm{Hom}}\nolimits (\Delta [1], M_*)_ n$ as a function $\Delta [1]_ n \to M_ n$. Using this notation, we define $h$ in degree $n$ by the rule

We first check $h$ is a morphism of cosimplicial modules. Namely, for $f : [n] \to [m]$ we will show that

The left hand side of (60.19.1.1) evaluated at $e_ i$ and then in turn evaluated at $\alpha ^ m_ j$ is

Note that $\alpha ^ m_ j \circ f = \alpha ^ n_{j'}$ where $0 \leq j' \leq n + 1$ is the unique index such that $f(i) < j$ if and only if $i < j'$. Thus the right hand side of (60.19.1.1) evaluated at $e_ i$ and then in turn evaluated at $\alpha ^ m_ j$ is

It follows from our description of $j'$ that the two answers are equal. Hence $h$ is a map of cosimplicial modules. Let $0 : \Delta [0] \to \Delta [1]$ and $1 : \Delta [0] \to \Delta [1]$ be the obvious maps, and denote $ev_0, ev_1 : \mathop{\mathrm{Hom}}\nolimits (\Delta [1], M_*) \to M_*$ the corresponding evaluation maps. The reader verifies readily that the compositions

are $0$ and $1$ respectively, whence $h$ is the desired homotopy between $0$ and $1$.

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