In general, if we have a map between CW-complexes, $f:X\to Y$, and $f$ is cellular, then is is clear that $f^{-1}(Y)$ (the inverse image, $f$ is not invertible in general) is also a CW-complex?
Since it's cellular, $f^{-1}(Y^n)$ must by closed and must contain $X^n$. I'm not sure what else I can say. Something about it intersecting a finite number of cells?
Ultimately, I am trying to use this to prove the following for CW-spectra: Let $f:E\to F$ be a function of spectra (in the strict sense here) and $F'$ be a cofinal subspectrum of $F$. Then there is a cofinal subspectrum $E'$ of $E$ such that $f$ maps $E'$ into $F'$.
My initial intuition was to show that $f^{-1}(F')$ was the desired subspectrum, but I got stymied at the very first step because I don't know much about CW spectra. I guess I could just work with nice spaces or something to simplify this...
Thanks! Jon