7
$\begingroup$

Let $R$ be a commutative ring spectrum, $M$ and $N$ be a $R$-module spectra.

Let us consider $R$-module maps from $M$ to $N$ up to stable homotopy, that is maps $M \to N$ such that the composites $R \wedge M \to M \xrightarrow{f} N$ and $R \wedge M \xrightarrow{1 \wedge f} R \wedge N \to N$ are equal in the stable homotopy category.

Now suppose that $M = R \wedge X$ is a free $R$-module.

Is it true that the set of stable homotopy classes of $R$-module maps up to homotopy in the previous sense, $[M, N]_R = [R \wedge X, N]_R$, is in the natural bijection with stable homotopy classes of all maps $[X, N]$?

I believe that something like that (or even stronger) holds for genuine $R$-module maps, but what about homotopy $R$-module maps if we are interested in the stable homotopy classes of maps only?

$\endgroup$
4
  • $\begingroup$ What do you mean by "homotopy $R$-module maps$? $\endgroup$ – Fernando Muro Aug 8 '20 at 16:07
  • $\begingroup$ Maps which are $R$-module maps only up to stable homotopy, as I said in the body of the question. The square diagram of the R-module condition is commutative up to stable homotopy, i. e. in stable homotopy category. The last question is just a less formal reformulation of the first. $\endgroup$ – Ann Aug 8 '20 at 16:34
  • 3
    $\begingroup$ This is true in every symmetric monoidal category. The Mai in one direction is composition with the unit of R and in the other tensoring with R followed by the action map of R on N. It is a routine to check they are inverse. $\endgroup$ – S. carmeli Aug 8 '20 at 19:27
  • $\begingroup$ Thank you very much! $\endgroup$ – Ann Aug 8 '20 at 20:06
4
$\begingroup$

In any symmetric monoidal category $C$, the free $R$-module functor $R\otimes -: C\to Mod_R(C)$ is left adjoint to the forgetful functor $U:Mod_R(C)\to C$. Hence, $Hom_{Mod_R(C)}(R\otimes X,N) \cong Hom_C(X,U(N))$. Since the stable homotopy category is symmetric monoidal, the result you asked for follows.

$\endgroup$
1
  • $\begingroup$ The question is ambiguous since it doesn't say whether $\wedge$ stands for the monoidal structure in the stable homotopy category or in a model. In the latter case, you'd need $X$ to be cofibrant so that both coincide. Another ambiguity is the use of "stable homotopy classes of maps". If it means morphisms in the stable homotopy category, no extra assumptions are needed, but if it means honest homotopy classes of maps (defined by means of cylinders or paths) you need in addition $N$ to be fibrant. $\endgroup$ – Fernando Muro Aug 11 '20 at 11:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.