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 $H\_\*(Y)$ ?



homology of a mapping spectrum
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. 


