Let $p$ be a prime number, let $k$ be a commutative ring in which $p=0$, and let $X = K( {\mathbb Z}/p {\mathbb Z}, n)$ be an Eilenberg-MacLane space. Let $F$ be the free $E_{\infty}$-algebra over $k$ generated by a class $\eta$ in (homological) degree $-n$. A result of Mandell asserts that there is a cofiber sequence of $E_{\infty}$-algebras over $k$ $$ F \stackrel{A}{\rightarrow} F \rightarrow C^{\ast}(X;k)$$ where $A$ is the ``Artin-Schreier'' map which carries $\eta$ to $\eta - P^0(\eta)$. In other words, as an $E_{\infty}$-algebra, the cochain complex $C^{\ast}(X;k)$ can be described by one generator (a class in degree $-n$) and one relation (the class should be fixed by $P^0$).

Is it possible to prove this result without explicitly computing the homotopy groups of the cofiber of $A$? Let's denote this cofiber by $R$. I can reduce to the problem of showing that $\pi_i R \simeq 0$ for $i > 0$ and that $\pi_0 R \simeq k$ (both of which are obvious consequences of Mandell's theorem). Is there some way to show this directly, without computing the other homotopy groups?


Bertrand Toen doesn't seem to do much calculation in Champs affines.

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    $\begingroup$ He also doesn't prove this theorem. Unless I misunderstand, he works in the setting of cosimplicial algebras (where the analogous statement is easy) and uses it to prove variants of Mandell's results. $\endgroup$ – Jacob Lurie Dec 3 '10 at 17:11
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    $\begingroup$ Toen's technique seems to involve an inductive approach, using the result for $X=K(Z/pZ,n)$ to prove it for $BX=K(Z/pZ,n+1)$. Could that be used here to reduce to the case of $n=1$ or $n=0$? $\endgroup$ – Charles Rezk Dec 3 '10 at 19:51
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    $\begingroup$ That's the strategy I had in mind. When n=0 you can prove it using deformation theory, so let's try induction on n. Let R(n) be the cofiber and let R'(n) be the cochains on K(Z/pZ,n). Then doing a bar construction on R(n) produces R(n-1), and similarly for R'(n). So the I.H. tells you that the map R(n) -> R'(n) is an equivalence after applying the bar construction. If you knew that R(n) had no positive homotopy and that pi_0 R(n) = k (statements which are obvious for R'(n)), then the bar construction doesn't lose any information and you are done. But a priori R(n) is a big mess. $\endgroup$ – Jacob Lurie Dec 3 '10 at 20:14

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