# homology of a mapping spectrum

If $X$ and $Y$ are two spectra, I denote by $F(X,Y)$ their mapping spectrum. This is uniquely determined by the existence of a natural isomorphism $[X\wedge Y, Z]\cong [X,F(Y,Z)]$.

I denote by $H_*$ the rational homology functor. If $X$ is a finite spectrum, I have an isomorphism $H_*(F(X,Y))\cong Hom(H_*(X),H_*(Y))$. Indeed, each side is a generalized cohomology theory in the $X$ variable that takes the value $H_*(Y)$ on the sphere. I would like to extend this result to a general $X$. What I can do is write a general spectrum $X$ as a filtered homotopy colimit of spectra $X_i$ that are finite. Then $F(X,Y)$ is the inverse limit of the spectra $F(X_i,Y)$. I get a Milnor exact sequence:

$$0\to \mathrm{lim}_i^1Hom_{n-1}(H_{*}(X_i),H_{*}(Y))\to H_nF(X,Y)\to \mathrm{lim}_i Hom_n(H_*(X_i),H_*(Y))\to 0$$

where $Hom_n$ denotes the set of homomorphisms of degree $n$.

I want to look at an example of the previous exact sequence with $X=\bigvee_{\mathbb{N}}S^0$ and $Y=S^0$. I can write $X$ as the homotopy colimit of $X_i=\bigvee_{1\leq k\leq i}S^0$. Then, $F(X,Y)$ is equivalent to $(S^0)^{\mathbb{N}}$, thus $H_0(F(X,Y))\cong \mathbb{Q}\otimes\mathbb{Z}^{\mathbb{N}}$. On the other hand, we have $$\mathrm{lim}_i\;Hom_0(H_*(X_i),H_*(Y))\cong \mathrm{lim}_i\;\mathbb{Q}^i\cong\mathbb{Q}^{\mathbb{N}}$$ The obvious map $\mathbb{Q}\otimes\mathbb{Z}^{\mathbb{N}}\to\mathbb{Q}^{\mathbb{N}}$ is not even surjective !

My question has two parts:

(1) Explain where the mistake is in the previous calculation. My guess is that I misunderstood the Milnor exact sequence.

(2) Is there a method for computing $H_*(F(X,Y))$ knowing $H_*(X)$ and $H_*(Y)$ ?

• @NeilStrickland. Thanks, this fully answers my question. If you want to write it as an answer, I will accept it. Are there easy to verify conditions on $X$ that insure that $H_*F(X,Y)\cong Hom(H_*X,H_*Y)$ ? Mar 19, 2015 at 11:11

This was a comment, but at the OP's suggestion I have promoted it to an answer.

There is no Milnor exact sequence for homology groups of an inverse limit, only for the homotopy groups. If $I$ is the Brown-Comenetz dual of the sphere, then $H_∗(I)=0$ but $F(I,I)$ is the profinite completion of S and so $H_∗F(I,I)≠0$. Thus, you cannot hope for a method that works in complete generality.

Since you are doing everything rationally and stably, homotopy and homology are the same.

So

$H_* F(X,Y) = [X,Y]_* \otimes \mathbb{Q}$

If $Y$ is itself a rational spectrum, then this is the same as

$\text{Hom}_{\mathbb{Q}}(H_* X, H_* Y)$

but of course in general it will not be. Said another way,

$\text{Hom}_{\mathbb{Q}}(H_* X, H_* Y) = H_*F(X,LY)$

where $LY$ is the rational localization of $Y$.