# On the natural (bigraded) homotopy groups of a simplicial object in a model category

$\def\mc{\mathcal} \def\sm{\wedge}$ This question stems from the Goerss-Hopkins paper Moduli Problems for Structured Ring Spectra. Let me begin by attempting to summarize the relevant framework -- this comes from the beginning of Chapter 3, on page 92.

Let $\mc{C}$ be a model category, and let $\mc{P}$ be a set of projectives defining the $\mc{P}$-resolution model stucture on $s\mc{C}$. (See Question 1 below.) Given $X \in s\mc{C}$ and for any $P\in \mc{P}$, define the $(n,P)$th natural homotopy group of $X$ by $\pi_{n,P}(X)=\pi_n(\mbox{map}(P,X))$ (using the derived mapping space). This is corepresentable in $\mbox{Ho}(s\mc{C})$ by $P \sm > \Delta^n/\partial\Delta^n$, which is defined as the pushout of the corner $\emptyset \otimes \ast \leftarrow P > \otimes \ast \rightarrow P \otimes > \Delta^n / \partial \Delta^n$. This smash product construction actually fits into an adjunction $$(-) \sm K : > \mc{C} / \emptyset \leftrightarrows s\mc{C} > : C_K,$$ where we define the right adjoint as follows. First, there is an adjunction $$(-)\otimes K:\mc{C} \leftrightarrows s\mc{C}:\mbox{hom}(K,-)_0 \stackrel{def}{=} > M_K(-)$$ (where the right adjoint is the $0$th object in the "hom object" of $s\mc{C}$ coming its simplicially contensored structure), and then we define $C_K$ via the pullback diagram $$\begin{array}{ccc} C_KX & > \rightarrow & M_KX \\ \downarrow & & > \downarrow \\ \emptyset & \rightarrow > & M_*X = X_0. \end{array}$$ In particular, write $C_nX=C_{\Delta^n / > \Lambda^n_0}X$ and $Z_nX=C_{\Delta^n / > \partial \Delta^n}X$. The "inclusion of the boundary" $\Delta^{n} > /\partial \Delta^{n} \rightarrow > \Delta^{n+1}/\Lambda^{n+1}_0$ induces the second map in a fibration sequence $$Z_{n+1} X \rightarrow C_{n+1}X \rightarrow > Z_{n}X$$ in the category $\mc{C}/\emptyset$ whenever $X \in s\mc{C}$ is Reedy fibrant. The neat result is that in this case, this in fact gives a "presentation" of the natural homotopy group $\pi_{n,P}(X)$ via the exact sequence $$[P,C_{n+1}X] \rightarrow > [P,Z_nX] \rightarrow \pi_{n,P}(X) > \rightarrow 0.$$

I have a few questions about the details and heuristics here.

Question 1: I'm confused about the extra conditions on $\mc{C}$ that might be necessary. They claim that the objects $P\in \mc{P}$ will be h-cogroup objects; this is what in particular gives a canonical map $P \rightarrow \emptyset$, which endows $\mbox{map}(P,X)$ with a basepoint. On the one hand, the definition of the $\mc{P}$-resolution model structure (see pages 24-26) starts out with the axiom that $\mc{P}$ be closed under suspension and desuspension, which suggests strongly that we're in a stable model category. But then it's silly to point out that the objects of $\mc{P}$ corepresent homotopy functors valued in abelian groups, since this is true of all objects. Even more, a stable model category is automatically pointed, which would mean that $\mc{C}/\emptyset$ is a totally vacuous thing to discuss; I'd consider this to be strong evidence that somehow $\mc{C}$ isn't meant to necessarily be stable. Of course the final application will be for $\mc{C}$ a category of spectra so this doesn't matter, but I've been pondering this long enough that the question has become of independent interest.

Question 2: Slightly further down (3.1.3 on page 95), they claim that the case $n=0$ of the result I cited yields an isomorphism $\pi_0([P,X]) \cong \pi_{0,P}(X)$ (where $[P,X]$ is a simplicial abelian group -- "simplicial" via $[P,X]_\bullet = [P,X_\bullet]$, and "abelian group" since supposedly $P$ is an (abelian, I guess) h-cogroup). However, unless I'm making a totally stupid mistake, I'm pretty sure that $Z_0X=C_{\Delta^0/\partial \Delta^0}X = C_*X = \emptyset$. I don't see how to make sense of this.

Question 3 (assuming $\mc{C}$ doesn't need to be stable): How should I understand taking a pullback over the initial object? If $\emptyset=*$, then I could consider $M_KX$ as "maps from $K$ to $X$" and $C_KX$ as "based maps from $K$ to $X$". This is really appealing; then I could view the "presentation" as saying that $\pi_{n,P}(X)$ is just $[P,-]$ applied to "based maps $S^n \rightarrow X$ mod based maps $D^{n+1} \rightarrow X$"! (It also makes sense that "based maps" should be adjoint to "smash product".) But as things stand, this isn't really honest. In fact, I don't think I've ever seen anyone take a pullback over an initial object (besides in the pointed case, of course). I can't seem to wrap my head around what $\mbox{lim}(\emptyset \rightarrow A \leftarrow B)$ should mean, besides "the last object $C$ over $B$ such that $C \rightarrow B \rightarrow A$ is trivial" (i.e. factors through $\emptyset$, obviously). This sounds an awful lot like the fiber of a map, but I'm not convinced that this analogy makes any sense. I'd be grateful to hear any geometric intuition that anyone can give here.

• Hi Aaron, I don't have time to write any long comments right now -- but if you are serious about looking for unstable results you should definitely take a look at the paper of Bousfield's that their resolution model structures are based on (arxiv.org/abs/math.AT/0312531) and dualize everything. – Tyler Lawson Oct 22 '12 at 1:08
• (Not to mention the Dwyer-Kan-Stover paper, of course.) – Tyler Lawson Oct 22 '12 at 1:09
• Thanks for the suggestion. I've already checked out Dwyer-Kan-Stover briefly, but it didn't seem to resolve these questions. And no, I don't care strongly about the unstable version -- my hands are already pretty full with trying to understand this paper! -- I've just become curious about it in the course of things. – Aaron Mazel-Gee Oct 22 '12 at 1:28

For question 2, my guess is that you're running into issues because $\Delta^0 / \partial \Delta^0$, the space obtained by taking a single point and collapsing the empty set to a single point, isn't $\ast$; by convention in homotopy theory, $X / \emptyset = X_+$.
• I see; thanks for the clarification. And yes, I suppose that convention $X/\emptyset = X_+$ makes sense: the rule is to take the subspace and identify it all with a (new) point. I'll leave this open for a few days in case anyone has anything to say about my third question. – Aaron Mazel-Gee Oct 24 '12 at 0:40