Let $X$ and $Y$ be CW complexes, and let $f$ and $g$ be homotopic cellular maps from $X$ to $Y$; that is, $f(X^n) \subset Y^n$ and $g(X^n) \subset Y^n$, where $X^n$ denotes the $n$-skeleton of $X$. How do I show that $f$ and $g$ are cellularly homotopic (homotopic via a homotopy that is itself a cellular map)?
My attempt. Consider the relative CW complex $(X \times I, X \times \partial I)$, and let $h: f \simeq g$. We may apply the cellular approximation theorem to $h:(X \times I, X \times \partial I) \to (Y,Y)$ to get a homotopy $H: h \simeq h' \text{ rel } X \times \partial I$ where $h': X\times I \to Y$ is cellular with $h'_0=f$ and $h'_1=g$.
But something is wrong—I have not used the fact that $f$ and $g$ are cellular!
EDIT (after contemplating freakish's answer):
I got confused originally partially because of the subtle statement of the cellular approximation theorem in May's A Concise Course in Algebraic Topology:
Theorem (Cellular Approximation). Any map $f: (X,A) \to (Y,B)$ between relative CW complexes is homotopic relative to $A$ to a cellular map.
My original understanding of this result was flawed; if I applied the result as stated above to my above attempt, what I would really obtain is a homotopy $H: h \simeq h' \text{ rel } X \times \partial I$ as above but a cellular map $h':(X \times I, X \times \partial I) \to (Y,Y)$ instead (with $h'_0=f$ and $h'_1=g$ as above). This is not the same as a cellular map $h': X \times I \to Y$, as is made obvious by the following wrong proof:
False result. All maps between CW complexes are cellular.
oof. Let $f:X \to Y$ be any map. Then we may view it as a map $f:(X,X) \to (Y,Y)$, so $f$ is homotopic relative to $X$ to a cellular map by cellular approximation. That is, the homotopy is constant on $X$, so that $f$ is cellular.
Though, in a similar light, any map $\varphi:(X,A) \to (Y,Y)$ is (trivially) cellular as $(Y,Y)^n=(Y,Y)^0=Y$ and $\varphi((X,A)^n) \subset Y$ trivially. Which means my original attempt was pretty flawed.
Here then is a proof that works, with much detail so that I (hopefully) will understand it in the future:
Proof that works. Let $f$, $g: X \to Y$ be homotopic cellular maps, and let $h: f \simeq g$. We wish to find a cellular homotopy $h': f \simeq g$; that is, a homotopy $h': X \times I \to Y$ between cellular maps that is a cellular map itself. That is, we require that $h'$ sends the $n$-skeleton $X^n \times \partial I \cup X^{n-1} \times I$ of $X \times I$ into $Y^n$. Since cellular homotopies are between cellular maps, $h'(X^n \times \partial I) \subset Y^n$ automatically, so it suffices to show that $h'(X^{n-1} \times I) \subset Y^n$.
Regard $h$ as a map of relative CW complexes $(X \times I, X^n \times \partial I) \to (Y, Y^n)$. Then the cellular approximation theorem gives us a homotopy $H: h \simeq h^n \text{ rel } X^n \times \partial I$ such that $h^n: (X \times I, X^n \times \partial I) \to (Y, Y^n)$ is cellular with $h^n_0|X^n=f|X^n$ and $h^n_1|X^n=g|X^n$.
Since $h^n$ is cellular, it takes the relative $n$-skeleton $X^n \times \partial I \cup X^{n-1} \times I$ of $(X \times I, X^n \times \partial I)$ into $Y^n$. Thus $h^n$ defines the desired homotopy $h'$ on $X^n$ for each $n \geq 1$, and we may take the colimit to obtain $h'$. $\square$
Though it is nicer to just use Hatcher's version of the cellular approximation theorem.
Best Answer
You did, it's just a hidden requirement of the cellular approximation theorem. Recall:
(see Allen Hatcher "Algebraic Topology", Theorem 4.8)
So in order to get that $h'_0=f$ and $h'_1=g$ you need to know that $H$ can be chosen to be stationary on $X\times \partial I$, which is a subcomplex. And this can be done if $h$ restricted to that subcomplex is cellular. And that requires $f$ and $g$ to be cellular.