# Homotopic cellular maps are cellularly homotopic

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.

## 1 Answer

But something is wrong—I have not used the fact that $$f$$ and $$g$$ are cellular!

You did, it's just a hidden requirement of the cellular approximation theorem. Recall:

Cellular Approximation Theorem: Every map $$f:X\to Y$$ of CW complexes is homotopic to a cellular map. If $$f$$ is already cellular on a subcomplex $$A\subseteq X$$ the homotopy may be taken to be stationary on $$A$$.

(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.