Injectivity of rationalization on spectrum morphisms

Question

Let $E$ and $F$ be two spectra, and let $j \colon F \to F_{\mathbb Q} = F \wedge H \mathbb Q$ be the rationalization of $F$.

Assume that the group of morphisms $[E, F]$ in the stable homotopy category is torsion free. Consider the natural map $j_* \colon [E, F] \to [E, F_{\mathbb Q}]$. If $E$ is finite, I think, $j_*$ is just the rationalization of the abelian group $[E, F]$, so is injective, since $[E, F]$ has no torsion.

Question. Is $j_*$ injective without the finiteness assumption on $E$ (if $[E, F]$ is torsion free)?

Is it true at least for bounded below $E$ and $F$ of finite type (i.e. with finitely generated homotopy groups)?

Answer 1

Another good example is to take $E=H$ and $F=\Sigma KU$. A key point here is that $KU\wedge H\simeq KU\wedge S\mathbb{Q}$. Indeed, the theory of Landweber exactness says that $\pi_*(KU\wedge H)$ is the universal example of a ring equipped with an isomorphism $f(x)$ from the additive formal group law to the multiplicative formal group law, which gives $\pi_*(KU\wedge H)=\mathbb{Q}[u,u^{-1}]$ with $f(x)=\exp(ux)-1$. From this we get a map $S\mathbb{Q}\to KU\wedge H$, which extends uniquely to a map $KU\wedge S\mathbb{Q}\to KU\wedge H$ of $KU$-modules, and this map is easily seen to be an equivalence. We therefore get $$ [H,\Sigma KU] = [KU\wedge H,\Sigma KU]^{KU} = [KU\wedge S\mathbb{Q},\Sigma KU]^{KU} = [S\mathbb{Q},\Sigma KU]. $$ We can now choose a free resolution $R\to F\to \mathbb{Q}$, giving a cofibration $SR\to SF\to S\mathbb{Q}$. By applying $[-,\Sigma KU]$ to this we get $[S\mathbb{Q},\Sigma KU]=\text{Ext}(\mathbb{Q},\mathbb{Z})$ (which is a rational vector space of uncountable dimension). On the other hand, we have $$ [H,\Sigma KU\mathbb{Q}]=\text{Hom}(\pi_*(H)\otimes\mathbb{Q},\pi_{*-1}(KU)\otimes\mathbb{Q}) = \text{Hom}(\mathbb{Q},\pi_{-1}(KU)\otimes\mathbb{Q})=0. $$

Here the homotopy groups of $F$ are finitely generated, but the homology groups are not, and $F$ is not connective. For a different kind of example, we could take $E=MU$ and $F=S^1$, so both $E$ and $F$ are $(-1)$-connected and of finite type. Using the Adams spectral sequence, Ravenel proved that $[MU,S/p]_*=0$ for all primes $p$. Using this one can deduce that $[MU\wedge S(\mathbb{Q}/\mathbb{Z}),S]_*=0$, and thus that $[MU,S^1]=[MU\mathbb{Q},S^1]$. Here $MU\mathbb{Q}$ is a wedge of even suspensions of $S\mathbb{Q}$, with $[S^{2n}\mathbb{Q},S^1]=0$ for $n>0$, and $[S\mathbb{Q},S^1]=\text{Ext}(\mathbb{Q},\mathbb{Z})$. It follows that once again we have $[E,F]=\text{Ext}(\mathbb{Q},\mathbb{Z})$ but $[E,F\mathbb{Q}]=0$.

Answer 2

My comment, with more details: No, $E = H\mathbb{Q}$ and $F = \Sigma\mathbb{Z}$ gives a counterexample. $H_0(E) = \mathbb{Q}$ and $H_i(E) = 0$ for $i \neq 0$, so the universal coefficient theorem gives $[E,F] = H^1(E) = \mathrm{Ext}^1_\mathbb{Z}(\mathbb{Q},\mathbb{Z})$, which is an uncountable $\mathbb{Q}$ vector space. In contrast, $[E,F_\mathbb{Q}] = \mathrm{Ext}^1_\mathbb{Z}(\mathbb{Q},\mathbb{Q}) = 0$, so $j_*$ is not injective in this case.

I don't think it suffices that $E$ and $F$ are bounded below and have finitely generated homotopy groups, either. For instance, take $E = \Sigma^\infty_+ BG$ for a finite group $G$ and take $F = ku$ the connective topological $K$-theory spectrum. Then $[E,F]$ is the completed representation ring of $G$, by the Atiyah--Segal completion theorem. For $G = \mathbb{Z}/2\mathbb{Z}$ it is additively isomorphic to $\mathbb{Z} \oplus \mathbb{Z}_2$, so it rationalizes to $\mathbb{Q} \oplus \mathbb{Q}_2$. In contrast, $$[E,F_\mathbb{Q}] = \prod_i H^{2i}(BG;\mathbb{Q}) = \mathbb{Q}.$$

Answer 3

As user171227 is pointing out, you are asking about the map $$ F^0(E) \rightarrow F\mathbb Q^0(E).$$ This always factors as $$ F^0(E) \rightarrow F^0(E) \otimes \mathbb Q \rightarrow F\mathbb Q^0(E).$$ The first map is injective if and only if $F^0(E)$ is torsion free, while the composite is injective if and only if the second map is also injective.

And happily, that second map often isn't, even when the first map is! It would have made the character theory that Mike Hopkins, Doug Ravenel, and I wrote about nonexistent, with $F$ equal to a Morava $E$-theory and $E = BG$. (This includes user17227's example: $K(BG)$ is always torsion free.)