I am trying to prove the following from Adams blue book (Lemma 2.6.(i)): Let $E$ and $F$ be CW-spectra, and $\tilde{F} \subseteq F$ be cofinal. Then there is a cofinal subspectrum $\tilde{E} \subseteq E$ with $f(\tilde{E} ) \subseteq \tilde{F}$.
My proof so far goes as follows: Let $\tilde{E}$ be the disjoint union over all subspectra $E^{'}$ of $E$ with $f(E^{'}) \subseteq \tilde{F}$. One can show that $\tilde{E}$ is a subspectra. For the cofinality, I need to show that $f^{-1}(\tilde{F}) \subseteq \tilde{E}$, and for this my idea was to show that $f^{-1}(\tilde{F})$ is in the set over which is taken the union for $\tilde{E}$, but for this I would need that preimages under cellular maps of CW-complexes are CW-complexes. Does anybody knows if this is true?
My understanding of your question is this:
In general this is not true. Here is an example.
Let $X = [0,1]$ with $0$-cells $\{0\}, \{1\}$ and $1$-cell $[0,1]$. Let $Y = S^1 = \{ z \in \mathbb C \mid \lvert z \rvert = 1\}$ with $0$-cell $\{1\}$ and $1$-cell $S^1$ (which is obtained by attaching $[0,1]$ to the $0$-skeleton $\{1\}$ via the constant map $\partial [0,1] \to \{1\}$).
Let $f : [0,1] \to S^1, f(t) = e^{4pi i t}$. This is a cellular map, but $f^{-1}(\{1\}) = \{0, 1/2, 1\}$ is not a subcomplex of $X$.