I'm reading Mike Hopkins' COCTALOS notes and having trouble with some pretty basic statements about (naive) spectra. Basically I'm nervous doing homological algebra with them, although in these problems I'm having I couldn't imagine the proofs being anything other than simple diagram chases. The relevant definitions are in section 4.

First, I'm having trouble with the proof of Prop. 4.12, which claims that every E-Adams resolution arises from an Adams tower. The proof is based on the diagram at the bottom of page 13 (sorry, I'm not sure how to put diagrams in an MO problem) and shows how to build the first level of the tower. To continue the tower, we need that two composite maps (at the top of page 14) are null, and this is what's confusing me. For (i), I have no idea, although $h$ must be E-mono since it's the first map in a factorization of an E-mono. For (ii), the problem is that we know both $\Sigma^{-1} I_0\rightarrow \Sigma^{-1}I_3$ and $\Sigma^{-1}I_1\rightarrow \Sigma^{-1}I_3$ are null, but the cofiber $X_1 = \Sigma^{-1}I_1/\Sigma^{-1}I_0$ may not map E-nully to $\Sigma^{-1}I_3$. In fact, there isn't necessarily a single choice of map anyways; it depends on the choice of nullhomotopy of $\Sigma^{-1}I_0\rightarrow \Sigma^{-1}I_3$ (factoring through $\Sigma^{-1}I_2$). All I know is that I need to find a map $C (E\wedge X_1) \simeq I\wedge E\wedge I_0 \rightarrow E\wedge \Sigma^{-1}I_3$ extending the map from $E\wedge X_1$. (Here $I$ is the pointed interval.)

Second, I don't understand the statement that "these maps are null after smashing with E and hence must actually be null as [their images] are E-injective". Suppose $A\rightarrow J$ is E-null and $J$ is E-injective. The only thing to do here is start with an E-mono $A\rightarrow B$ and obtain a lift $B\rightarrow J$ in the diagram $B\leftarrow A \rightarrow J$. Then the lift will have to be E-null too. But I don't see how this would imply that $A\rightarrow J$ is actually null. And there's no obvious E-mono I've got to work with anyways, besides maybe ${Id}_A$ (which doesn't help at all).

And lastly, I'm having trouble proving the $\Leftarrow$ direction of Lemma 5.2, that if for $E$ a ring spectrum the map $J=S\wedge J\rightarrow E\wedge J$ is the inclusion of a retract then $J$ is E-injective. I'd prove this by taking an E-mono $X\rightarrow Y$ and trying to find a lift in $Y\leftarrow X \rightarrow J$ for any $X\rightarrow J$. The only thing to do here is smash with $E$ and obtain $E\wedge B\leftarrow E\wedge A \rightarrow E\wedge J \rightarrow J$ (where the first map is mono), but I can't see what this buys me. There's one fact we've got about monos (Lemma 4.3), but not sure whether it helps here: $A \rightarrow B$ is mono iff there is some $C\rightarrow B$ such that $A\vee C\rightarrow B$ is a weak equivalence.

Since these seem like they should be such routine verifications, I'm feeling that there must be some basic technique I'm missing. Does anyone have any suggestions?

  • $\begingroup$ It's probably not directly relevant to your questions, but you may want to look at Haynes Miller's notes about Hopkins' proof of the exact functor theorem. Miller's notes are based on the ones you reference, and are available here: www-math.mit.edu/~hrm/papers/bcat.pdf $\endgroup$
    – Dan Ramras
    Jun 9 '11 at 19:01

Suppose that the inclusion $\eta\wedge 1:I\to E\wedge I$ admits a retraction, say $r$. Consider an $E$-monomorphism $f:A\to B$ and an arbitrary map $g:A\to I$. As $f$ is an $E$-monomorphism we can choose a retraction $s:E\wedge B\to E\wedge A$, and then the composite $$ h = (B \xrightarrow{\eta\wedge 1} E\wedge B \xrightarrow{s} E\wedge A \xrightarrow{1\wedge g} E\wedge I \xrightarrow{r} I) $$ satisfies $hf=g$. This shows that $I$ is $E$-injective.

Now suppose we have an $E$-null map $f:A\to I$, where $I$ is $E$-injective. This gives a commutative square

\begin{array}{ccccc} & A & \xrightarrow{f} & I & \\ \eta\wedge 1 & \downarrow & & \downarrow &\eta\wedge 1 \\ & E\wedge A & \xrightarrow{f\wedge 1} & E\wedge I \end{array}

The bottom map is zero, so if we go down then across we get zero, so if we go across and then down we also get zero. However, the right hand vertical map is a split monomorphism, so we can conclude that $f$ itself is zero.

This covers your second and third questions. I'm not going to discuss the first one because it is too painful to do the diagrams, but hopefully the other two will give you some ideas.

  • $\begingroup$ Thanks a lot. Your answer to my third question (the first of your two answers) is exactly what I was looking for, but the other claim was made (at the top of page 14) before any assumptions about $E$ being a ring spectrum. Perhaps there's another way to do it? $\endgroup$ Jun 12 '11 at 19:17

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