Stable homotopy groups commute with inverse limit

Question

Suppose we have a family of spectra $(E_i)_{i \in I}$ such that the inverse limit $\lim_i E_i$ does exist in the stable homotopy category (i.e. $\lim_i E_i$ is the limit in $\mathrm{SHC}$, the stable homotopy category). I want to show that $\pi_n(\lim_i E_i) \cong \lim_i \pi_n(E_i)$ for $n \in \mathbb{N}$. My proof is as follows: It is $\pi_n(E) = [\Sigma^n \mathbb{S}, E]$ for a spectrum $E$, where $\mathbb{S}$ is the sphere spectrum, and $[-, -]$ are the morphism sets in the stable homotopy category. We know that the Hom-functor commutes with inverse limits in the second variable. It is now \begin{align*} \pi_n(\lim_i E_i) = [\Sigma^n \mathbb{S}, \lim_i E_i] = \mathrm{Hom}_{\mathrm{SHC}}(\Sigma^n \mathbb{S}, \lim_i E_i ) \cong \lim_i \mathrm{Hom}_{\mathrm{SHC}}(\Sigma^n, E_i) = \lim_i [\Sigma^n \mathbb{S}, E_i] = \lim_i \pi_n(E_i). \end{align*} Is this proof correct? It seems a bit too easy for me, especially since the ordinary limits dont commute with the homotopy groups.

Answer 1

Limits tend not to exist in the stable homotopy category, just weak limits: the difference is that the universal property of a limit is that a particular map exists and is unique, whereas in a weak limit, you just know that the particular map exists. If we write $E$ for a weak limit of the inverse system $\{E_i\}$, then for any $Y$, the map $[Y, E] \to \varprojlim [Y, E_i]$ is surjective, but it may not be injective: Hom doesn't automatically convert a weak limit to a limit. For example given the inverse system $$ \dots \to E_2 \to E_1 \to E_0, $$ there is a Milnor exact sequence $$ 0 \to \lim{}^1 [Y, E_i] \to [Y, E] \to \lim [Y, E_i] \to 0. $$