Confusion about Phantom Maps A phantom cohomology operation (originally read phantom map) $f: X \rightarrow Y$ is a non-nullhomotopic map such that the induced cohomology operation on the cohomology theories for spaces is trivial. It is known that phantom maps exist.
Could someone tell me where my argument for the opposite fails?
Suppose a map $f:X \rightarrow Y$ is a phantom map and that $X,Y$ are represented by sequential CW $\Omega$-spectra and $f$ by a map of sequential spectra. Since $f$ is nontrivial, there must be some $f_n :X_n \rightarrow Y_n$ on which it is non-nullhomotopic (if I had to guess, I'd presume this is where the error lies). Then this implies that $Y^n (Y_n)$ detects the non-triviality of $f$ because $[Y_n,Y_n] \ni 1 \rightarrow f_n$ under the cohomology operation associated to $f$.
If I am right about the error, could anyone provide intuition about why a map of spectra can have all its components nullhomotopic without being trivial?

I'll translate my argument to the language of suspension spectra which makes it more clear where the error lies:
Let's suppose I decompose $X$ as a colimit of suspension spectra. Then this phantom operation $f$ has the property that the restriction of $f$ to any of the objects in the colimit is trivial. This is precisely the statement that $ f \in \operatorname{Ker}([\operatorname{colim}X_i,Y]) \rightarrow \operatorname{lim} [X_i ,Y]$. This is precisely what the $\operatorname{lim}^1$ term describes. In general, cohomology of a colimit is not the direct limit of the cohomologies.
I believe this is the same phenomenon behind the fact that a map of spectra (as I alluded to in my question) can have all its components nullhomotopic without being nullhomotopic itself. This is similar to the fact that maps between CW complexes exist which have their restriction to all finite CW complexes nullhomotopic without the entire map being so.
It is perhaps worth mentioning that none of these issues arise with homology in terms of bringing homotopy limits outside of homology.
