Let $A$ be a connected $E_{n+1}$-ring spectrum and let $\alpha\in\pi_k(A)$. I am having trouble showing that attaching an $E_n$-cell along $\alpha$ will necessarily not kill an element $\beta\in\pi_k(A)$ unless $\beta$ is a multiple of $\alpha$. Perhaps this is not true, but it seems plausible to me since attaching an $E_n$-cell seems like everything above the cell that kills $\alpha$ itself should be in higher dimensions.

To be precise: we can "attach an $E_n$-$A$-cell" to $A$ along $\alpha$ by taking the pushout of the following span in $E_n$-$A$-algebras: $$A\overset{\overline{0}}\leftarrow F_{E_n}(\Sigma^kA)\overset{\overline{\alpha}}\to A $$ where $F_{E_n}$ is the free $E_n$-$A$-algebra functor, $\overline{0}$ is the adjoint of the zero map $\Sigma^kA\to A$ and $\overline{\alpha}$ is the adjoint of the $A$-module map $\Sigma^kA\to A$ induced by $\alpha$.

It seems to me that there should be a "bottom layer" of $F_{E_n}(\Sigma^kA)$ that kills of $\alpha$, but that everything else (used to kill off the powers of $\alpha$ in a homotopy coherent way) should happen in higher dimensions, so cannot kill $\beta$. Is this true?

It may be useful to notice that, according to Lemma 4.4 of this paper of Antolin-Camarena and Barthel, that the above pushout is equivalent to $Ind_0^n(cof(\Sigma^k A\to A))$ where $Ind_0^n$ is the left adjoint to the forgetful functor from $E_n$-$A$-algebras to $E_0$-$A$-algebras (where $E_0$-$A$-algebras here just means unital $A$-modules).

I should also mention that this is the pretty clearly the best we can do in general, since as Tyler pointed out in the comments it's relatively easy to attach a structured cell along something in degree $k$ and kill something in degree $k+1$. And his example is not rare either. I can construct an infinite family of ring spectra in which this occurs for arbitrarily large degrees, when only attaching $E_1$-$A$-cells.

  • $\begingroup$ Unfortunately not. For example, I'd expect that killing off $x$ also kills off brackets like $\langle x,y,z \rangle$ which may not be multiples of $x$ in homotopy. As another example, killing off 2 in the sphere spectrum in an $E_1$ way also kills the Hopf map $\eta$. $\endgroup$ Jul 19 '17 at 13:03
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    $\begingroup$ @TylerLawson: I think he asking the following: if $t \in \pi_k(A)$ maps to zero in $\pi_k(A//\alpha)$, is it then the case that $t = a_0 \alpha$, for some $a_0 \in \pi_0(A)$. (Note that $t$ and $\alpha$ live in the same degree.) Thus $\eta$ is not a counterexample, not sure if brackets might be. $\endgroup$ Jul 19 '17 at 13:48
  • $\begingroup$ Sorry I'm having a hard time explaining this clearly. But I'm only really interested in things which are in the same degree as $\alpha$. So $\eta$ is not an issue. $\endgroup$ Jul 19 '17 at 15:12
  • $\begingroup$ Yeah so to be clear, since $\beta$ is also in $\pi_k$, it must be "multiples" by elements in $\pi_0$. $\endgroup$ Jul 19 '17 at 17:59
  • $\begingroup$ I'm sorry, Jon, I didn't read carefully enough. $\endgroup$ Jul 20 '17 at 1:04

[I write $A//\alpha$ for the pushout computed in $E_n-A$-algebras, and $E/\alpha$ for the pushout computed in $A$-modules.]

Here is an attempt (for $k>0$), though I should make it clear that I consider myself an amateur at this. I number all my claims so misstakes are easier to point out :).

(1) The forgetful functor $E_n-A-Alg \to E_n-Alg$ has a left adjoint given by smashing with $A$.

(2) All forgetful functors in sight preserve sifted colimits.

(3) The pushout $A//\alpha$ can be computed as geometric realisation of a two-sided bar construction $|Bar_{F_{E_n}(\Sigma^k A)}(A, A)|$ This is a sifted colimit. We have $B_n := Bar_{F_{E_n}(\Sigma^k A)}(A, A)_n = A \vee_A [F_{E_n}(\Sigma^k A)]^{\vee_A n} \vee_A A$, where $\vee_A$ means the coproduct in $E_n-A$-algebras.

(4) The free algebra functors are left adjoint, so preserve colimits. It follows from (1) that $A \vee_A [F_{E_n}(\Sigma^k A)]^{\vee_A n} \vee_A A = G_{E_n}[(S^k)^{\vee n}]\wedge A$, where $G_{E_n}$ denotes the free $E_n-S$-algebra functor.

(5) We have $G_{E_n}(X) = \bigvee_{k \ge 0} E_n(k)_+ \times_{\Sigma_k}^h X^{\wedge k}$.

(6) Thus $B_n = E_n(0)_+ \wedge A \vee E_n(1)_+ \wedge [(S^k)^{\vee n}] \wedge A \vee B_n'$, where $B_n'$ is $2k$-connective and so $k$-connected. (This is where I use $k>0$.)

(7) Since $E_n(0)$ and $E_n(1)$ are contractible, we have $B_n = A \vee (\Sigma^k A)^{\vee n} \vee B_n'$. We thus get the following decomposition of $A//\alpha$ (as a cofiber sequence): $$ |C_\bullet| \to A //\alpha \to |B'_\bullet|. $$ Here $C_n = A \vee (\Sigma^k A)^{\vee n} \vee 0 = Bar_{\Sigma^k A}(A, 0)$ and so $|C_\bullet|$ is the cofiber of $\alpha$. On the other hand each $B'_n$ is $k$-connected and hence so is $|B'_\bullet|$. It follows that $\pi_i(A//\alpha) = \pi_i(A/\alpha)$ for $i \le k$.

edit: I did not check that the decompositions I write down are compatible with the simplicial structure maps. But note that there is a map $A/\alpha \to A//\alpha$ under the identity on $A$. To construct it, using that $F_{E_n}$ preserves colimits, by adjunction it suffices to factor $\bar{\alpha}: F_{E_n} \Sigma^k A \to A$ through $F_{E_n} \alpha$. But by definition $\bar{\alpha}$ is $F_{E_n} \Sigma^k A \xrightarrow{F_{E_n} \alpha} F_{E_n} A \xrightarrow{\eta} A$, where $\eta$ is the (co)unit of adjunction. Surely this must be the map I write down...

  • $\begingroup$ When you write $A/ \alpha$, what does it mean? Is it just the spectrum cofiber of $\mathrm S \xrightarrow{\alpha} A$? Because that can't be right, at the very least you must factor out all $\pi_0(A)\alpha$. $\endgroup$ Jul 19 '17 at 15:45
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    $\begingroup$ I mean the cofiber of $\Sigma^n A \to A$, which I suppose actually is $S^n \wedge A \to A \wedge A \to A$, where the last map is multiplication. $\endgroup$ Jul 19 '17 at 16:02
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    $\begingroup$ Well things can be killed in degree 2k and higher, right?, but I suppose the last line can be improved to "for $i<2k$" $\endgroup$ Jul 20 '17 at 5:29
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    $\begingroup$ @JonathanBeardsley I have edited my post $\endgroup$ Jul 21 '17 at 6:04
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    $\begingroup$ I wrote an "answer" that's really just long comments on your answer, with a strategy to define the simplicial maps. $\endgroup$ Jul 26 '17 at 4:48

This is not an answer but rather extended commentary on Tom Bachmann's answer. I hope to address some of Jon's comments on that answer.

Bar constructions with respect to coproducts compute pushouts

First let me address Tom's point (3), that an instance of the two-sided bar construction computes pushouts. This is probably well-known but I had to think about it for a bit, so I'll include an explanation in case it helps someone else. Also, this explanation helps later with coherence issues for maps between simplicial objects.

The two-sided bar construction can be defined very generally, for a monoid $M$ for some monoidal structure $\otimes$ together with left and right $M$-modules $L$ and $R$. Now consider the special case when $\otimes$ is given by the coproduct. In this case, the monoidal structure on $M$ is unique, given by the fold $M \sqcup M \to M$, and the module structures on $L$ and $R$ are given simply by morphisms $M \to L$, $M \to R$. And the claim is that the geometric realization of the bar construction is simply the pushout of $R \leftarrow M \to L$.

One way to see that is to consider the functor $b : \mathrm{Span} \to \Delta^\mathrm{op}$ where $\mathrm{Span} = \{ r \leftarrow m \to l \}$ and $b(m \to l) = (d_0 : [1] \to [0])$ and $b(m \to r) = (d_1 : [1] \to [0])$. It's not hard to check that the coproduct-based bar construction $\mathrm{Fun}(\mathrm{Span}, \mathcal{C}) \to \mathrm{Fun}(\Delta^\mathrm{op}, \mathcal{C})$ is given by left Kan extension along this functor $b$. The usual argument about left Kan extending along the composite $\mathrm{Span} \to \Delta^\mathrm{op} \to \ast$, shows that a span and its corresponding bar construction have the same colimit.

Defining the relevant map of simplicial $A$-modules

Instead of trying to check that Tom's maps are compatible with faces and degeneracies, let's try to build simplicial maps wholesale. I'll use $F : \mathrm{Mod}_A \to \mathrm{Alg}^{E_n}_{A}$ for the free $E_n$-$A$-algebra functor defined on $A$-module spectra, and $U : \mathrm{Alg}^{E_n}_{A} \to \mathrm{Mod}_A$ for the corresponding forgetful functor.

The key ingredient in Tom's argument is a simplicial $A$-module $B_{\bullet}$ whose geometric realization is $U(A/\!/\alpha)$, the underlying $A$-module of the $E_n$-$A$-algebra in Jon's question. By the previous section, the simplicial $E_n$-$A$-algebra $\mathcal{B}_{\bullet} := \mathrm{Lan}_b(A\overset{\overline{0}}\leftarrow {E_n}(\Sigma^kA)\overset{\overline{\alpha}}\to A)$ has geometric realization given by the pushout of that span, $A /\!/ \alpha$. Since the forgetful functor $U$ preserves geometric realizations, the simplicial $A$-module $B_{\bullet} := U \circ \mathcal{B}_{\bullet}$ will have $U(A /\!/ \alpha)$ as geometric realization.

Now let $C_{\bullet} := \mathrm{Lan}_b(0 \leftarrow \Sigma^k A \to A)$ be the coproduct-based bar construction in $\mathrm{Mod}_A$. We want to define a morphism of simplicial $A$-modules $C_{\bullet} \to B_{\bullet} = U \circ \mathcal{B}_{\bullet}$, or equivalently a morphism of simplicial $E_n$-$A$-algebras $F \circ C_{\bullet} \to \mathcal{B}_{\bullet}$. Now, since $F$ is a left adjoint it preserves the left Kan extension defining $C_{\bullet}$, so $F \circ C_{\bullet}$ is the bar construction for the span $F(0) \leftarrow F(\Sigma^k A) \to F(A)$. The following diagram is a natural transformation between that span and the one defining $\mathcal{B}_{\bullet}\require{AMScd}$: $$\begin{CD} F(0) @<{F(0)}<< F(\Sigma^k A) @>{F(\alpha)}>> F(A) \\ @V{\mathrm{id}}VV @V{\mathrm{id}}VV @V{\mu_A}VV \\ A @<{\bar{0}}<< F(\Sigma^k A) @>{\bar{\alpha}}>> A \\ \end{CD}$$ This map of spans induces a simplicial map between their respective bar constructions.

The rest of the argument

I think that now the rest of Tom's argument runs fine: let $B'_{\bullet}$ be the cofibre in simplicial $A$-modules of $C_{\bullet} \to B_{\bullet}$. Now, without worrying about compatibility with faces and degeneracies, you can identify for each $n$ the map $C_n \to B_n$ to describe $B'_n$ and see that $B'_n$ is $2k$-connective.

  • $\begingroup$ Can you or @Tom say why the map you construct here is the same map that picks out the relevant pieces in Tom's decomposition? It doesn't seem obvious to me. $\endgroup$ Sep 22 '17 at 7:37

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