Lemma 60.19.2. With notation as in (60.17.0.5) the complex

is homotopic to zero as a $D(*)$-cosimplicial module.

Lemma 60.19.2. With notation as in (60.17.0.5) the complex

\[ \Omega _{D(0)} \to \Omega _{D(1)} \to \Omega _{D(2)} \to \ldots \]

is homotopic to zero as a $D(*)$-cosimplicial module.

**Proof.**
We are going to use the principle of Simplicial, Lemma 14.28.4 and more specifically Lemma 60.16.1 which tells us that homotopic maps between (co)simplicial objects are transformed by any functor into homotopic maps. The complex of the lemma is equal to the $p$-adic completion of the base change of the cosimplicial module

\[ M_* = \left( \Omega _{P/A} \to \Omega _{P \otimes _ A P/A} \to \Omega _{P \otimes _ A P \otimes _ A P/A} \to \ldots \right) \]

via the cosimplicial ring map $P\otimes _ A \ldots \otimes _ A P \to D(n)$. This follows from Lemma 60.6.6, see comments following (60.17.0.2). Hence it suffices to show that the cosimplicial module $M_*$ is homotopic to zero (uses base change and $p$-adic completion). We can even assume $A = \mathbf{Z}$ and $P = \mathbf{Z}[\{ x_ i\} _{i \in I}]$ as we can use base change with $\mathbf{Z} \to A$. In this case $P^{\otimes n + 1}$ is the polynomial algebra on the elements

\[ x_ i(e) = 1 \otimes \ldots \otimes x_ i \otimes \ldots \otimes 1 \]

with $x_ i$ in the $e$th slot. The modules of the complex are free on the generators $\text{d}x_ i(e)$. Note that if $f : [n] \to [m]$ is a map then we see that

\[ M_*(f)(\text{d}x_ i(e)) = \text{d}x_ i(f(e)) \]

Hence we see that $M_*$ is a direct sum over $I$ of copies of the module studied in Example 60.19.1 and we win. $\square$

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## Comments (1)

Comment #8713 by oyen4d on