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Let $M$ be a model category and $G$ a finite group, and equip the category $M^G$ of $G$-objects in $M$ with, say, a projective model structure. Then there is a canonical functor

$$\mathrm{Ho}(M^G) \to \mathrm{Ho}(M)^G. $$

Surprisingly (to me), if $M$ is additive and the cardinality of $G$ is invertible in $M$, then this functor is fully faithful.

I learned this from 1303.0153 which says that although it is "certainly well-known, we haven’t been able to find it in the literature". But surely someone has noticed this or something like it before? Even an offhand remark without a proof would be a useful reference, even in some special case like chain complexes over $\mathbb{Q}$ or localized spectra.

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  • $\begingroup$ This may be relevant: mathematik.uni-bielefeld.de/sfb343/preprints/pr98032.ps.gz $\endgroup$
    – Ben Wieland
    Commented Jun 6, 2014 at 3:21
  • $\begingroup$ @BenWieland hmm, can you explain why? $\endgroup$
    – Mike Shulman
    Commented Jun 6, 2014 at 15:46
  • $\begingroup$ Cooke's paper ("Replacing homotopy actions by topological actions", TransAMS) proves this for M=p-local spaces (e.g., Theorem 2.3). $\endgroup$
    – Charles Rezk
    Commented Jul 28, 2014 at 22:11
  • $\begingroup$ @CharlesRezk Thanks! That's exactly the sort of paper that I felt sure must exist. If you post that as an answer, I'll accept it. $\endgroup$
    – Mike Shulman
    Commented Jul 30, 2014 at 4:25
  • $\begingroup$ @CharlesRezk hmm, although now that I look at it more carefully, his Theorem 2.3 actually shows that the functor in question is essentially surjective, rather than fully faithful! $\endgroup$
    – Mike Shulman
    Commented Jul 31, 2014 at 22:41

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The spectral sequence I constructed with Niles Johnson was precisely designed to handle questions of this sort (here is a version that is closer to the publication version: T-algebra SS). A special case of your question is considered in Section 5.1. Our methods require a suitably enriched model category (we focus on simplicial model categories), but it is easy to see that you get a similar spectral sequence for spectral or chain complex enrichments. In such a case the obstructions to the existence and uniqueness of a lift of a given map $f\colon X\rightarrow Y$ from $\mathrm{Ho}(M)^G$ to $\mathrm{Ho}(M^G)$ lie in the positive dimensional Borel cohomology of $X$ with coefficients in various shifts of $Y$. Using a tensor and cotensor with spaces, we can write this more precisely: the obstructions are the $t$th cohomology groups of the cosimplicial abelian group $\mathrm{ho}(M^G)_{\downarrow Y}(G^{\bullet+1}\otimes X,Y^{S^t}))$ for $t>0$. Here the source has the simplicial structure coming from a bar construction.

When the order of $G$ acts invertibly on these cohomology groups, the restriction and transfer homomorphisms exhibit these groups as retracts of the corresponding cohomology with the trivial group. Of course these groups are trivial in positive degrees so there is a unique lift of each map.

Of course, I am assuming some kind of nice enriched model structure here, so I am not answering your question exactly. On the other hand, in these case I think you can get away with less: You probably just need $Y$ to be a homotopy monoid in $\mathrm{Ho}(M)$ (to get the $E_2$ description as a bunch of abelian groups) and you only need the order of $G$ to act invertibly on this $E_2$ term in positive cohomological degrees.

I apologize that the linked reference is not quite complete. We are in the middle of making revisions for publication. The article will later appear in Advances in Mathematics.

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  • $\begingroup$ Thanks! So in particular, if $M$ is additive, then every object is a homotopy monoid, so the result should apply. It's interesting that in the special case of actions by a group with invertible order one can prove it without such an involved calculation. $\endgroup$
    – Mike Shulman
    Commented Jun 6, 2014 at 15:47
  • $\begingroup$ As often happens, behind a vanishing spectral sequence argument lies a more elementary argument. My more involved method is probably overkill in this case. Nonetheless, the argument indicates why it is true. For $X,Y\in M^G$ we have a map $$ \mathrm{tr}\colon M(X,Y)\rightarrow M^G(X,Y)$$ given by $f(-)\mapsto \sum_{g\in G} (g\cdot f)(-)=\sum_{g\in G} gf(g^{-1} -)$. I suspect that $\mathrm{tr}$ descends to homotopy categories, perhaps by an argument with right homotopies. If this is the case, then precomposing with the restriction functor gives a self map on mapping sets in $\mathrm{Ho}(M^G)$.. $\endgroup$
    – Justin Noel
    Commented Jun 6, 2014 at 18:41
  • $\begingroup$ . whose composite is multiplication by $|G|$ and which factors through $\mathrm{Ho}(M)^G(X,Y)$. From here one can see that $\mathrm{Ho}(M^G)(X,Y)\cong\mathrm{Ho}(M)^G(X,Y)$ when $|G|$ acts invertibly. It appears the non-trivial part is seeing that $\mathrm{tr}$ descends to the homotopy category. $\endgroup$
    – Justin Noel
    Commented Jun 6, 2014 at 18:50

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