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I have a probably very basic question about modules over symmetric ring spectra:

Let $R$ be a commutative symmetric ring spectrum and let $M$ and $N$ be module spectra over $R$. Moreover, let $\varphi \colon M \to N$ be a morphism of $R$-module spectra, which is a stable equivalence of the underlying symmetric spectra. Moreover, denote by $End_R(M)$ and $End_R(N)$ the spectra of $R$-linear endomorphisms (using the inner hom spectra). If there is any justice in the world the answer to the following should be "yes":

Is $End_R(N)$ stably equivalent to $End_R(M)$?

This question can be reduced to the following one: Are the maps $\varphi_* \colon End_R(M) \to Hom_R(M,N)$ and $\varphi^* \colon End_R(N) \to Hom_R(M,N)$ induced by pre- and postcomposition stable equivalences?

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    $\begingroup$ If N and M are fibrant-cofibrant R-Modules then the answer is yes. $\endgroup$
    – Ilias A.
    Commented Jul 2, 2014 at 19:02
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    $\begingroup$ As @Fedotov suggests, derived mapping objects are homotopy invariant in any enriched model category, that means that mapping objects of fibrant-cofibrant objects are invariant. You can otherwise construct counterexamples, e.g. in chain complexes over a ring, the plain endomorphism ring of a module is concentrated in degree 0, while the derived endomorphism DGA has (in general) nontrivial higher homology: the Ext algebra. $\endgroup$
    – Fernando Muro
    Commented Jul 2, 2014 at 22:28
  • $\begingroup$ So, if $N$ and $M$ are fibrant-cofibrant, then $End_R(N)$ is stably equivalent to $End_R(M)$ as symmetric spectra. Are they also equivalent as ring spectra? $\endgroup$
    – Ulrich Pennig
    Commented Aug 1, 2014 at 9:59

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Yes, this is true. This kind of property for a model category is a consequence of what is sometimes abusively called the "SM7" axiom for the enrichment:

  • In $R$-modules, suppose that $A \to B$ is a cofibration and $C \to D$ is a fibration. Then the natural map of $R$-modules $$ Hom_R(B,C) \to Hom_R(A,C) \times_{Hom_R(A,D)} Hom_R(B,D) $$ is a fibration of $R$-modules, which is a weak equivalence if either of the original maps is.

(If the enrichment were, instead, an enrichment in simplicial sets, then this would genuinely be Quillen's SM7 axiom for simplicial model categories.)

Let's suppose for the time being that $N \to M$ is a cofibration and a weak equivalence, $N$ is cofibrant, and $P$ is an arbitrary fibrant $R$-module. Applying the SM7 axiom to $N \to M$ and $P \to *$ shows that $Hom_R(M,P) \to Hom_R(N,P)$ is (a fibration and) a weak equivalence.

This shows $Hom_R(-,P)$ takes acyclic cofibrations between cofibrant objects to equivalences; Ken Brown's lemma then implies that it takes arbitrary weak equivalences between cofibrant objects to equivalences.

A dual proof shows that, if $Q$ is cofibrant, $Hom_R(Q,-)$ takes weak equivalences between fibrant objects to equivalences.

When $N$ and $M$ are cofibrant-fibrant, we then find that all relevant maps of Hom-objects induced by pre- or post-composition with $\phi$ are equivalences.

Side note: There's a subtlety here if you're using one of the "positive" model structures of Shipley's "A convenient model category for commutative ring spectra" (called the S-model structure there). There, in order for things to be homotopically well-behaved, you have to interpret the positive model structure as actually enriched in the ordinary stable model structure (Shipley proves this version of the SM7 axiom and more).

Further side note: You didn't ask, but you might worry about whether you can soup this up to an equivalence of $R$-algebras. You can, but only up to a "zigzag" weak equivalence (which can't be gotten around in general) by using the endomorphism object of the map $\phi$ -- I learned this technique from Rezk's thesis. (For enrichments in spaces, an earlier paper of Dwyer and Kan gives a much more powerful tool for constructing derived endomorphism spaces.) A proof that you get equivalent endomorphism algebras (which also provides functoriality in weak equivalences, but only in the homotopy category of algebras) can be found in Section 2 of this paper.

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    $\begingroup$ Geez, I need to learn more model category theory. Thank you! $\endgroup$
    – Ulrich Pennig
    Commented Jul 3, 2014 at 7:37
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    $\begingroup$ Considering my last comment, it actually sounded quite negative. Actually, I am looking forward to learning more about model categories :-). $\endgroup$
    – Ulrich Pennig
    Commented Jul 3, 2014 at 11:56
  • $\begingroup$ If pre- and postcomposition with $\phi$ are weak equivalences, then the endomorphism spectra are equivalent as (symmetric) spectra. Are they also equivalent as ring spectra? $\endgroup$
    – Ulrich Pennig
    Commented Aug 1, 2014 at 10:04
  • $\begingroup$ @UlrichPennig Neither $Hom_R(M,N)$ nor $Hom_R(N,M)$ has a strictly associative ring structure, and one of these is always the source or target of such a composition operation. As a result, to construct an equivalence between $End_R(M)$ and $End_R(N)$ you really do need a third object with a ring structure to move between them, as in the third paragraph. (If you allow yourself to talk about $A_\infty$ algebras instead, the answer to your question is "yes".) However, if you just work in the derived category, then yes: all four admit ring spectrum structures compatible with $\phi$. $\endgroup$
    – Tyler Lawson
    Commented Aug 2, 2014 at 5:43

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