Is the pre-image of a cellular map a CW complex?

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

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Perhaps a proof by induction? That's how things always seem to be proved about CW complexes... – Jon Beardsley Oct 5 '11 at 14:38

I guess you are reading Adams' book?

So in Adams' terms a (strict) function $f$ of degree $r$ is a sequence of maps $f_n:E_n \to F_{n-r}$ such that the relevant diagram commutes for all $n \in \mathbb{Z}$

So we simply set $E'$ to be the subspectrum of cells of the $E_n$ that are mapped into $E'_{n-r}$. You can then check that this forms a cofinal spectrum.

(Actually I found this difficult as well. Either someone told me how to do it, or I found a proof somewhere, so really I shouldn't take credit for this)

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Thankyou! Joyfully, I figured this out while waiting on the curb for my ride today! But this confirms the exact solution I discovered! The notions about CW complexes end up being irrelevant. – Jon Beardsley Oct 6 '11 at 2:29
In fact, the question is ill posed, as I should be asking about the preimage of a subcomplex... Not the whole complex. Does Adams assume that the maps are cellular, or is this not necessary, as he doesn't seem to state it explicitly, but I use that fact in my proof. – Jon Beardsley Oct 6 '11 at 2:30
I think all maps can assumed to be cellular by cellular approximation – Juan S Oct 6 '11 at 3:36
Sure, I mean, we might as well assume that, as long as we don't care about things up to homotopy, right? I was just wondering if there was some reason Adams never explicitly said that. – Jon Beardsley Oct 6 '11 at 15:25